柳婼の blog – 第192页 – 我不管，反正我最萌～

L1-005. 考试座位号-PAT团体程序设计天梯赛GPLT

每个PAT考生在参加考试时都会被分配两个座位号，一个是试机座位，一个是考试座位。正常情况下，考生在入场时先得到试机座位号码，入座进入试机状态后，系统会显示该考生的考试座位号码，考试时考生需要换到考试座位就座。但有些考生迟到了，试机已经结束，他们只能拿着领到的试机座位号码求助于你，从后台查出他们的考试座位号码。

输入格式：

输入第一行给出一个正整数N（<=1000），随后N行，每行给出一个考生的信息：“准考证号试机座位号考试座位号”。其中准考证号由14位数字组成，座位从1到N编号。输入保证每个人的准考证号都不同，并且任何时候都不会把两个人分配到同一个座位上。

考生信息之后，给出一个正整数M（<=N），随后一行中给出M个待查询的试机座位号码，以空格分隔。

输出格式：

对应每个需要查询的试机座位号码，在一行中输出对应考生的准考证号和考试座位号码，中间用1个空格分隔。

输入样例：
4
10120150912233 2 4
10120150912119 4 1
10120150912126 1 3
10120150912002 3 2
2
3 4
输出样例：
10120150912002 2
10120150912119 1

#include <iostream>
#include <vector>
using namespace std;
struct stu {
    string s;
    int one, two;
};
int main() {
    int n, m, temp;
    cin >> n;
    vector<stu> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i].s >> a[i].one >> a[i].two;
    }
    cin >> m;
    for (int i = 0; i < m; i++) {
        cin >> temp;
        for (int j = 0; j < n; j++) {
            if (a[j].one == temp) {
                cout << a[j].s << " " << a[j].two << endl;
                break;
            }
        }
    }
    return 0;
}

#include <iostream>

#include <vector>

using namespace std;

struct stu {

string s;

int one, two;

};

int main() {

int n, m, temp;

cin >> n;

vector<stu> a(n);

for (int i = 0; i < n; i++) {

cin >> a[i].s >> a[i].one >> a[i].two;

}

cin >> m;

for (int i = 0; i < m; i++) {

cin >> temp;

for (int j = 0; j < n; j++) {

if (a[j].one == temp) {

cout << a[j].s << " " << a[j].two << endl;

break;

}

return 0;

}

L1-004. 计算摄氏温度-PAT团体程序设计天梯赛GPLT

给定一个华氏温度F，本题要求编写程序，计算对应的摄氏温度C。计算公式：C = 5*(F-32)/9。题目保证输入与输出均在整型范围内。

输入格式：
输入在一行中给出一个华氏温度。

输出格式：
在一行中按照格式“Celsius = C”输出对应的摄氏温度C的整数值。

输入样例：
150
输出样例：
Celsius = 65

#include <iostream>
using namespace std;
int main() {
    int f;
    cin >> f;
    int c = 5 * (f - 32) / 9;
    cout << "Celsius = " << c;
    return 0;
}

#include <iostream>

using namespace std;

int main() {

int f;

cin >> f;

int c = 5 * (f - 32) / 9;

cout << "Celsius = " << c;

return 0;

}

L1-002. 打印沙漏-PAT团体程序设计天梯赛GPLT

本题要求你写个程序把给定的符号打印成沙漏的形状。例如给定17个“*”，要求按下列格式打印

*****
***
*
***
*****
所谓“沙漏形状”，是指每行输出奇数个符号；各行符号中心对齐；相邻两行符号数差2；符号数先从大到小顺序递减到1，再从小到大顺序递增；首尾符号数相等。

给定任意N个符号，不一定能正好组成一个沙漏。要求打印出的沙漏能用掉尽可能多的符号。

输入格式：
输入在一行给出1个正整数N（<=1000）和一个符号，中间以空格分隔。

输出格式：
首先打印出由给定符号组成的最大的沙漏形状，最后在一行中输出剩下没用掉的符号数。

输入样例：
19 *
输出样例：
*****
***
*
***
*****
2

#include <iostream>
using namespace std;
int main() {
    int N, row = 0;
    char c;
    cin >> N >> c;
    for (int i = 0; i < N; i++) {
        if ((2 * i * (i + 2) + 1) > N) {
            row = i - 1;
            break;
        }
    }
    for (int i = row; i >= 1; i--) {
        for (int k = row - i; k >= 1; k--) cout << " ";
        for (int j = i * 2 + 1; j >= 1; j--) cout << c;
        cout << endl;
    }
    for (int i = 0; i < row; i++) cout << " ";
    cout << c << endl;
    for (int i = 1; i <= row; i++) {
        for (int k = row - i; k >= 1; k--) cout << " ";
        for (int j = i * 2 + 1; j >= 1; j--) cout << c;
        cout << endl;
    }
    cout << (N - (2 * row * (row + 2) + 1));
    return 0;
}

#include <iostream>

using namespace std;

int main() {

int N, row = 0;

char c;

cin >> N >> c;

for (int i = 0; i < N; i++) {

if ((2 * i * (i + 2) + 1) > N) {

row = i - 1;

break;

}

for (int i = row; i >= 1; i--) {

for (int k = row - i; k >= 1; k--) cout << " ";

for (int j = i * 2 + 1; j >= 1; j--) cout << c;

cout << endl;

}

for (int i = 0; i < row; i++) cout << " ";

cout << c << endl;

for (int i = 1; i <= row; i++) {

for (int k = row - i; k >= 1; k--) cout << " ";

for (int j = i * 2 + 1; j >= 1; j--) cout << c;

cout << endl;

}

cout << (N - (2 * row * (row + 2) + 1));

return 0;

}

L1-001. Hello World-PAT团体程序设计天梯赛GPLT

这道超级简单的题目没有任何输入。

你只需要在一行中输出著名短句“Hello World!”就可以了。

#include <iostream>
using namespace std;
int main() {
    cout << "Hello World!";
    return 0;
}

#include <iostream>

using namespace std;

int main() {

cout << "Hello World!";

return 0;

}

L3-007. 天梯地图-PAT团体程序设计天梯赛GPLT

本题要求你实现一个天梯赛专属在线地图，队员输入自己学校所在地和赛场地点后，该地图应该推荐两条路线：一条是最快到达路线；一条是最短距离的路线。题目保证对任意的查询请求，地图上都至少存在一条可达路线。

输入格式：

输入在第一行给出两个正整数N（2 <= N <=500）和M，分别为地图中所有标记地点的个数和连接地点的道路条数。随后M行，每行按如下格式给出一条道路的信息：

V1 V2 one-way length time

其中V1和V2是道路的两个端点的编号（从0到N-1）；如果该道路是从V1到V2的单行线，则one-way为1，否则为0；length是道路的长度；time是通过该路所需要的时间。最后给出一对起点和终点的编号。

输出格式：

首先按下列格式输出最快到达的时间T和用节点编号表示的路线：

Time = T: 起点 => 节点1 => … => 终点

然后在下一行按下列格式输出最短距离D和用节点编号表示的路线：

Distance = D: 起点 => 节点1 => … => 终点

如果最快到达路线不唯一，则输出几条最快路线中最短的那条，题目保证这条路线是唯一的。而如果最短距离的路线不唯一，则输出途径节点数最少的那条，题目保证这条路线是唯一的。

如果这两条路线是完全一样的，则按下列格式输出：

Time = T; Distance = D: 起点 => 节点1 => … => 终点

输入样例1：
10 15
0 1 0 1 1
8 0 0 1 1
4 8 1 1 1
5 4 0 2 3
5 9 1 1 4
0 6 0 1 1
7 3 1 1 2
8 3 1 1 2
2 5 0 2 2
2 1 1 1 1
1 5 0 1 3
1 4 0 1 1
9 7 1 1 3
3 1 0 2 5
6 3 1 2 1
5 3
输出样例1：
Time = 6: 5 => 4 => 8 => 3
Distance = 3: 5 => 1 => 3
输入样例2:
7 9
0 4 1 1 1
1 6 1 3 1
2 6 1 1 1
2 5 1 2 2
3 0 0 1 1
3 1 1 3 1
3 2 1 2 1
4 5 0 2 2
6 5 1 2 1
3 5
输出样例2:
Time = 3; Distance = 4: 3 => 2 => 5
分析：用两个Dijkstra + DFS。一个求最快路径（如果相同求路径的那条），一个求最短路径（如果相同求结点数最小的那条）～～求最快路径可以直接在Dijkstra里面求前驱结点Timepre数组~~~求最短路径因为要求结点数最小的那条，所以要用dispre的二维数组存储所有结点的最短路径，然后用DFS求出满足条件的结点数最小的那条～～

注意：
1.一开始最后一个测试用例“答案错误”，后来发现是自己在求最短路径（第二个答案distance）的时候忘记了temppath每一次深搜结束后的pop_back();
2.如果直接使用DFS的话，会导致最后一个测试用例“运行超时”～

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = 999999999;
int dis[510], Time[510], e[510][510], w[510][510], Timepre[510], weight[510];
bool visit[510];
vector<int> Timepath, dispath, temppath, dispre[510];
int st, fin, minnode = inf;
void dfsTimepath(int v) {
    Timepath.push_back(v);
    if(v == st) {
        return ;
    }
    dfsTimepath(Timepre[v]);
}
void dfsdispath(int v) {
    temppath.push_back(v);
    if(v == st) {
        if(temppath.size() < minnode) {
            minnode = temppath.size();
            dispath = temppath;
        }
        temppath.pop_back();
        return ;
    }
    for(int i = 0; i < dispre[v].size(); i++) {
        dfsdispath(dispre[v][i]);
    }
    temppath.pop_back();
}
int main() {
    fill(dis, dis + 510, inf);
    fill(Time, Time + 510, inf);
    fill(weight, weight + 510, inf);
    fill(e[0], e[0] + 510 * 510, inf);
    fill(w[0], w[0] + 510 * 510, inf);
    int n, m;
    scanf("%d %d", &n, &m);
    int a, b, flag, len, t;
    for(int i = 0; i < m; i++) {
        scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);
        e[a][b] = len;
        w[a][b] = t;
        if(flag != 1) {
            e[b][a] = len;
            w[b][a] = t;
        }
    }
    scanf("%d %d", &st, &fin);
    Time[st] = 0;
    for(int i = 0; i < n; i++) {
        Timepre[i] = i;
    }
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && Time[j] < minn) {
                u = j;
                minn = Time[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v < n; v++) {
            if(visit[v] == false && w[u][v] != inf) {
                if(w[u][v] + Time[u] < Time[v]) {
                    Time[v] = w[u][v] + Time[u];
                    Timepre[v] = u;
                    weight[v] = weight[u] + e[u][v];
                } else if(w[u][v] + Time[u] == Time[v] && weight[v] > weight[u] + e[u][v]) {
                    weight[v] = weight[u] + e[u][v];
                    Timepre[v] = u;
                }
            }
        }
    }
    dfsTimepath(fin);
    fill(visit, visit + 510, false);
    dis[st] = 0;
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && minn > dis[j]) {
                u = j;
                minn = dis[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v < n; v++) {
            if(visit[v] == false && e[u][v] != inf) {
                if(e[u][v] + dis[u] < dis[v]) {
                    dis[v] = e[u][v] + dis[u];
                    dispre[v].clear();
                    dispre[v].push_back(u);
                } else if(e[u][v] + dis[u] == dis[v]) {
                    dispre[v].push_back(u);
                }
            }
        }
    }
    dfsdispath(fin);
    printf("Time = %d", Time[fin]);
    if(dispath == Timepath) {
        printf("; Distance = %d: ", dis[fin]);
    } else {
        printf(": ");
        for(int i = Timepath.size() - 1; i >= 0; i--) {
            printf("%d", Timepath[i]);
            if(i != 0) printf(" => ");
        }
        printf("\n");
        printf("Distance = %d: ", dis[fin]);
    }
    for(int i = dispath.size() - 1; i >= 0; i--) {
        printf("%d", dispath[i]);
        if(i != 0) printf(" => ");
    }
    return 0;
}

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#include <cstdio>

#include <algorithm>

#include <vector>

using namespace std;

const int inf = 999999999;

int dis[510], Time[510], e[510][510], w[510][510], Timepre[510], weight[510];

bool visit[510];

vector<int> Timepath, dispath, temppath, dispre[510];

int st, fin, minnode = inf;

void dfsTimepath(int v) {

Timepath.push_back(v);

if(v == st) {

return ;

}

dfsTimepath(Timepre[v]);

}

void dfsdispath(int v) {

temppath.push_back(v);

if(v == st) {

if(temppath.size() < minnode) {

minnode = temppath.size();

dispath = temppath;

}

temppath.pop_back();

return ;

}

for(int i = 0; i < dispre[v].size(); i++) {

dfsdispath(dispre[v][i]);

}

temppath.pop_back();

}

int main() {

fill(dis, dis + 510, inf);

fill(Time, Time + 510, inf);

fill(weight, weight + 510, inf);

fill(e[0], e[0] + 510 * 510, inf);

fill(w[0], w[0] + 510 * 510, inf);

int n, m;

scanf("%d %d", &n, &m);

int a, b, flag, len, t;

for(int i = 0; i < m; i++) {

scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);

e[a][b] = len;

w[a][b] = t;

if(flag != 1) {

e[b][a] = len;

w[b][a] = t;

}

scanf("%d %d", &st, &fin);

Time[st] = 0;

for(int i = 0; i < n; i++) {

Timepre[i] = i;

}

for(int i = 0; i < n; i++) {

int u = -1, minn = inf;

for(int j = 0; j < n; j++) {

if(visit[j] == false && Time[j] < minn) {

u = j;

minn = Time[j];

}

if(u == -1) break;

visit[u] = true;

for(int v = 0; v < n; v++) {

if(visit[v] == false && w[u][v] != inf) {

if(w[u][v] + Time[u] < Time[v]) {

Time[v] = w[u][v] + Time[u];

Timepre[v] = u;

weight[v] = weight[u] + e[u][v];

} else if(w[u][v] + Time[u] == Time[v] && weight[v] > weight[u] + e[u][v]) {

weight[v] = weight[u] + e[u][v];

Timepre[v] = u;

}

dfsTimepath(fin);

fill(visit, visit + 510, false);

dis[st] = 0;

for(int i = 0; i < n; i++) {

int u = -1, minn = inf;

for(int j = 0; j < n; j++) {

if(visit[j] == false && minn > dis[j]) {

u = j;

minn = dis[j];

}

if(u == -1) break;

visit[u] = true;

for(int v = 0; v < n; v++) {

if(visit[v] == false && e[u][v] != inf) {

if(e[u][v] + dis[u] < dis[v]) {

dis[v] = e[u][v] + dis[u];

dispre[v].clear();

dispre[v].push_back(u);

} else if(e[u][v] + dis[u] == dis[v]) {

dispre[v].push_back(u);

}

dfsdispath(fin);

printf("Time = %d", Time[fin]);

if(dispath == Timepath) {

printf("; Distance = %d: ", dis[fin]);

} else {

printf(": ");

for(int i = Timepath.size() - 1; i >= 0; i--) {

printf("%d", Timepath[i]);

if(i != 0) printf(" => ");

}

printf("\n");

printf("Distance = %d: ", dis[fin]);

}

for(int i = dispath.size() - 1; i >= 0; i--) {

printf("%d", dispath[i]);

if(i != 0) printf(" => ");

}

return 0;

}

1111. Online Map (30)-PAT甲级真题（Dijkstra + DFS）

Input our current position and a destination, an online map can recommend several paths. Now your job is to recommend two paths to your user: one is the shortest, and the other is the fastest. It is guaranteed that a path exists for any request.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (2 <= N <= 500), and M, being the total number of streets intersections on a map, and the number of streets, respectively. Then M lines follow, each describes a street in the format:

V1 V2 one-way length time

where V1 and V2 are the indices (from 0 to N-1) of the two ends of the street; one-way is 1 if the street is one-way from V1 to V2, or 0 if not; length is the length of the street; and time is the time taken to pass the street.

Finally a pair of source and destination is given.

Output Specification:

For each case, first print the shortest path from the source to the destination with distance D in the format:

Distance = D: source -> v1 -> … -> destination

Then in the next line print the fastest path with total time T:

Time = T: source -> w1 -> … -> destination

In case the shortest path is not unique, output the fastest one among the shortest paths, which is guaranteed to be unique. In case the fastest path is not unique, output the one that passes through the fewest intersections, which is guaranteed to be unique.

In case the shortest and the fastest paths are identical, print them in one line in the format:

Distance = D; Time = T: source -> u1 -> … -> destination

Sample Input 1:
10 15
0 1 0 1 1
8 0 0 1 1
4 8 1 1 1
3 4 0 3 2
3 9 1 4 1
0 6 0 1 1
7 5 1 2 1
8 5 1 2 1
2 3 0 2 2
2 1 1 1 1
1 3 0 3 1
1 4 0 1 1
9 7 1 3 1
5 1 0 5 2
6 5 1 1 2
3 5
Sample Output 1:
Distance = 6: 3 -> 4 -> 8 -> 5
Time = 3: 3 -> 1 -> 5
Sample Input 2:
7 9
0 4 1 1 1
1 6 1 1 3
2 6 1 1 1
2 5 1 2 2
3 0 0 1 1
3 1 1 1 3
3 2 1 1 2
4 5 0 2 2
6 5 1 1 2
3 5
Sample Output 2:
Distance = 3; Time = 4: 3 -> 2 -> 5

题目大意：给一张地图，两个结点中既有距离也有时间，有的单行有的双向，要求根据地图推荐两条路线：一条是最快到达路线，一条是最短距离的路线。
第一行给出两个整数N和M，表示地图中地点的个数和路径的条数。接下来的M行每一行给出：道路结点编号V1 道路结点编号V2 是否单行线道路长度所需时间
要求第一行输出最快到达时间Time和路径，第二行输出最短距离Distance和路径

分析：
1.用两个Dijkstra。一个求最短路径（如果相同求时间最短的那条），一个求最快路径（如果相同求结点数最小的那条）～～
2.求最短路径,和最快路径都可以在Dijkstra里面求前驱结点dispre和，Timepre数组～
3.dispre数组更新的条件是路径更短，或者路径相等的同时时间更短。
4.求最快路径时候要多维护一个NodeNum数组，记录在时间最短的情况下，到达此节点所需的节点数量。
Time数组更新的条件是，时间更短，时间相同的时候，如果此节点能让到达次节点是数目也变小，则更新Timepre，heNodeNum数组
5.最后根据dispre 和Timepre数组递归出两条路径，比较判断，输出最终答案～
注意：如果直接使用DFS的话，会导致最后一个测试用例“运行超时”～～

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = 999999999;
int dis[510], Time[510], e[510][510], w[510][510], dispre[510],Timepre[510], weight[510],NodeNum[510];
bool visit[510];
vector<int> dispath, Timepath, temppath;
int st, fin, minnode = inf;
void dfsdispath(int v) {
    dispath.push_back(v);
    if(v == st) return ;
    dfsdispath(dispre[v]);
}
void dfsTimepath(int v) {
    Timepath.push_back(v);
    if(v == st) return ;
    dfsTimepath(Timepre[v]);
}
int main() {
    fill(dis, dis + 510, inf);
    fill(Time, Time + 510, inf);
    fill(weight, weight + 510, inf);
    fill(e[0], e[0] + 510 * 510, inf);
    fill(w[0], w[0] + 510 * 510, inf);
    int n, m;
    scanf("%d %d", &n, &m);
    int a, b, flag, len, t;
    for(int i = 0; i < m; i++) {
        scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);
        e[a][b] = len;
        w[a][b] = t;
        if(flag != 1) {
            e[b][a] = len;
            w[b][a] = t;
        }
    }
    scanf("%d %d", &st, &fin);
    dis[st] = 0;
    for(int i = 0; i < n; i++)
        dispre[i] = i;
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && dis[j] < minn) {
                u = j;
                minn = dis[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v < n; v++) {
            if(visit[v] == false && e[u][v] != inf) {
                if(e[u][v] + dis[u] < dis[v]) {
                    dis[v] = e[u][v] + dis[u];
                    dispre[v] = u;
                    weight[v] = weight[u] + w[u][v];
                } else if(e[u][v] + dis[u] == dis[v] && weight[v] > weight[u] + w[u][v]) {
                    weight[v] = weight[u] + w[u][v];
                    dispre[v] = u;
                }
            }
        }
    }
    dfsdispath(fin);
    Time[st] = 0;
    fill(visit, visit + 510, false);
    for(int i = 0; i < n; i++) {
        int u = -1, minn = inf;
        for(int j = 0; j < n; j++) {
            if(visit[j] == false && minn > Time[j]) {
                u = j;
                minn = Time[j];
            }
        }
        if(u == -1) break;
        visit[u] = true;
        for(int v = 0; v < n; v++) {
            if(visit[v] == false && w[u][v] != inf) {
                if(w[u][v] + Time[u] < Time[v]) {
                    Time[v] = w[u][v] + Time[u];
                    Timepre[v]= u;
                    NodeNum[v]=NodeNum[u]+1;
                } else if(w[u][v] + Time[u] == Time[v]&&NodeNum[u]+1<NodeNum[v]) {
                    Timepre[v]= u;
                    NodeNum[v]=NodeNum[u]+1;
                }
            }
        }
    }
    dfsTimepath(fin);
    printf("Distance = %d", dis[fin]);
    if(dispath == Timepath) {
        printf("; Time = %d: ", Time[fin]);
    } else {
        printf(": ");
        for(int i = dispath.size() - 1; i >= 0; i--) {
            printf("%d", dispath[i]);
            if(i != 0) printf(" -> ");
        }
        printf("\nTime = %d: ", Time[fin]);
    }
    for(int i = Timepath.size() - 1; i >= 0; i--) {
        printf("%d", Timepath[i]);
        if(i != 0) printf(" -> ");
    }
    return 0;
}

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#include <iostream>

#include <algorithm>

#include <vector>

using namespace std;

const int inf = 999999999;

int dis[510], Time[510], e[510][510], w[510][510], dispre[510],Timepre[510], weight[510],NodeNum[510];

bool visit[510];

vector<int> dispath, Timepath, temppath;

int st, fin, minnode = inf;

void dfsdispath(int v) {

dispath.push_back(v);

if(v == st) return ;

dfsdispath(dispre[v]);

}

void dfsTimepath(int v) {

Timepath.push_back(v);

if(v == st) return ;

dfsTimepath(Timepre[v]);

}

int main() {

fill(dis, dis + 510, inf);

fill(Time, Time + 510, inf);

fill(weight, weight + 510, inf);

fill(e[0], e[0] + 510 * 510, inf);

fill(w[0], w[0] + 510 * 510, inf);

int n, m;

scanf("%d %d", &n, &m);

int a, b, flag, len, t;

for(int i = 0; i < m; i++) {

scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);

e[a][b] = len;

w[a][b] = t;

if(flag != 1) {

e[b][a] = len;

w[b][a] = t;

}

scanf("%d %d", &st, &fin);

dis[st] = 0;

for(int i = 0; i < n; i++)

dispre[i] = i;

for(int i = 0; i < n; i++) {

int u = -1, minn = inf;

for(int j = 0; j < n; j++) {

if(visit[j] == false && dis[j] < minn) {

u = j;

minn = dis[j];

}

if(u == -1) break;

visit[u] = true;

for(int v = 0; v < n; v++) {

if(visit[v] == false && e[u][v] != inf) {

if(e[u][v] + dis[u] < dis[v]) {

dis[v] = e[u][v] + dis[u];

dispre[v] = u;

weight[v] = weight[u] + w[u][v];

} else if(e[u][v] + dis[u] == dis[v] && weight[v] > weight[u] + w[u][v]) {

weight[v] = weight[u] + w[u][v];

dispre[v] = u;

}

dfsdispath(fin);

Time[st] = 0;

fill(visit, visit + 510, false);

for(int i = 0; i < n; i++) {

int u = -1, minn = inf;

for(int j = 0; j < n; j++) {

if(visit[j] == false && minn > Time[j]) {

u = j;

minn = Time[j];

}

if(u == -1) break;

visit[u] = true;

for(int v = 0; v < n; v++) {

if(visit[v] == false && w[u][v] != inf) {

if(w[u][v] + Time[u] < Time[v]) {

Time[v] = w[u][v] + Time[u];

Timepre[v]= u;

NodeNum[v]=NodeNum[u]+1;

} else if(w[u][v] + Time[u] == Time[v]&&NodeNum[u]+1<NodeNum[v]) {

Timepre[v]= u;

NodeNum[v]=NodeNum[u]+1;

}

dfsTimepath(fin);

printf("Distance = %d", dis[fin]);

if(dispath == Timepath) {

printf("; Time = %d: ", Time[fin]);

} else {

printf(": ");

for(int i = dispath.size() - 1; i >= 0; i--) {

printf("%d", dispath[i]);

if(i != 0) printf(" -> ");

}

printf("\nTime = %d: ", Time[fin]);

}

for(int i = Timepath.size() - 1; i >= 0; i--) {

printf("%d", Timepath[i]);

if(i != 0) printf(" -> ");

}

return 0;

}