Code – 第194页

1021. Deepest Root (25)-PAT甲级真题（图的遍历，dfs，连通分量的个数）

A graph which is connected and acyclic can be considered a tree. The height of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N-1 lines follow, each describes an edge by given the two adjacent nodes’ numbers.

Output Specification:

For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print “Error: K components” where K is the number of connected components in the graph.

Sample Input 1:
5
1 2
1 3
1 4
2 5
Sample Output 1:
3
4
5
Sample Input 2:
5
1 3
1 4
2 5
3 4
Sample Output 2:
Error: 2 components
题目大意：给出n个结点（1~n）之间的n条边，问是否能构成一棵树，如果不能构成则输出它有的连通分量个数，如果能构成一棵树，输出能构成最深的树的高度时，树的根结点。如果有多个，按照从小到大输出。
分析：首先深度优先搜索判断它有几个连通分量。如果有多个，那就输出Error: x components，如果只有一个，就两次深度优先搜索，先从一个结点dfs后保留最高高度拥有的结点们，然后从这些结点中的其中任意一个开始dfs得到最高高度的结点们，这两个结点集合的并集就是所求

#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
int n, maxheight = 0;
vector<vector<int>> v;
bool visit[10010];
set<int> s;
vector<int> temp;
void dfs(int node, int height) {
    if(height > maxheight) {
        temp.clear();
        temp.push_back(node);
        maxheight = height;
    } else if(height == maxheight){
        temp.push_back(node);
    }
    visit[node] = true;
    for(int i = 0; i < v[node].size(); i++) {
        if(visit[v[node][i]] == false)
            dfs(v[node][i], height + 1);
    }
}
int main() {
    scanf("%d", &n);
    v.resize(n + 1);
    int a, b, cnt = 0, s1 = 0;
    for(int i = 0; i < n - 1; i++) {
        scanf("%d%d", &a, &b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    for(int i = 1; i <= n; i++) {
        if(visit[i] == false) {
            dfs(i, 1);
            if(i == 1) {
                if (temp.size() != 0) s1 = temp[0];
                for(int j = 0; j < temp.size(); j++)
                    s.insert(temp[j]);
            }
            cnt++;
        }
    }
    if(cnt >= 2) {
        printf("Error: %d components", cnt);
    } else {
        temp.clear();
        maxheight = 0;
        fill(visit, visit + 10010, false);
        dfs(s1, 1);
        for(int i = 0; i < temp.size(); i++)
            s.insert(temp[i]);
        for(auto it = s.begin(); it != s.end(); it++)
            printf("%d\n", *it);
    }
    return 0;
}

#include <iostream>

#include <vector>

#include <set>

#include <algorithm>

using namespace std;

int n, maxheight = 0;

vector<vector<int>> v;

bool visit[10010];

set<int> s;

vector<int> temp;

void dfs(int node, int height) {

if(height > maxheight) {

temp.clear();

temp.push_back(node);

maxheight = height;

} else if(height == maxheight){

temp.push_back(node);

}

visit[node] = true;

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

if(visit[v[node][i]] == false)

dfs(v[node][i], height + 1);

}

int main() {

scanf("%d", &n);

v.resize(n + 1);

int a, b, cnt = 0, s1 = 0;

for(int i = 0; i < n - 1; i++) {

scanf("%d%d", &a, &b);

v[a].push_back(b);

v[b].push_back(a);

}

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

if(visit[i] == false) {

dfs(i, 1);

if(i == 1) {

if (temp.size() != 0) s1 = temp[0];

for(int j = 0; j < temp.size(); j++)

s.insert(temp[j]);

}

cnt++;

}

if(cnt >= 2) {

printf("Error: %d components", cnt);

} else {

temp.clear();

maxheight = 0;

fill(visit, visit + 10010, false);

dfs(s1, 1);

for(int i = 0; i < temp.size(); i++)

s.insert(temp[i]);

for(auto it = s.begin(); it != s.end(); it++)

printf("%d\n", *it);

}

return 0;

}

1013. Battle Over Cities (25)-PAT甲级真题（图的遍历，统计强连通分量的个数，dfs）

It is vitally important to have all the cities connected by highways in a war. If a city is occupied by the enemy, all the highways from/toward that city are closed. We must know immediately if we need to repair any other highways to keep the rest of the cities connected. Given the map of cities which have all the remaining highways marked, you are supposed to tell the number of highways need to be repaired, quickly.

For example, if we have 3 cities and 2 highways connecting city1-city2 and city1-city3. Then if city1 is occupied by the enemy, we must have 1 highway repaired, that is the highway city2-city3.

Input

Each input file contains one test case. Each case starts with a line containing 3 numbers N (<1000), M and K, which are the total number of cities, the number of remaining highways, and the number of cities to be checked, respectively. Then M lines follow, each describes a highway by 2 integers, which are the numbers of the cities the highway connects. The cities are numbered from 1 to N. Finally there is a line containing K numbers, which represent the cities we concern.

Output

For each of the K cities, output in a line the number of highways need to be repaired if that city is lost.

Sample Input
3 2 3
1 2
1 3
1 2 3
Sample Output
1
0
0

题目大意：给出n个城市之间有相互连接的m条道路，当删除一个城市和其连接的道路的时候，问其他几个剩余的城市至少要添加多少个路线才能让它们重新变为连通图
分析：添加的最少的路线，就是他们的连通分量数-1，因为当a个互相分立的连通分量需要变为连通图的时候，只需要添加a-1个路线，就能让他们相连。所以这道题就是求去除了某个结点之后其他的图所拥有的连通分量数
使用邻接矩阵存储，对于每一个被占领的城市，去除这个城市结点，就是把它标记为已经访问过，这样在深度优先遍历的时候，对于所有未访问的结点进行遍历，就能求到所有的连通分量的个数～记得因为有很多个要判断的数据，每一次输入被占领的城市之前要把visit数组置为false～～

#include <cstdio>
#include <algorithm>
using namespace std;
int v[1010][1010];
bool visit[1010];
int n;
void dfs(int node) {
    visit[node] = true;
    for(int i = 1; i <= n; i++) {
        if(visit[i] == false && v[node][i] == 1)
            dfs(i);
    }
}
int main() {
    int m, k, a, b;
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 0; i < m; i++) {
        scanf("%d%d", &a, &b);
        v[a][b] = v[b][a] = 1;
    }
    for(int i = 0; i < k; i++) {
        fill(visit, visit + 1010, false);
        scanf("%d", &a);
        int cnt = 0;
        visit[a] = true;
        for(int j = 1; j <= n; j++) {
            if(visit[j] == false) {
                dfs(j);
                cnt++;
            }
        }
        printf("%d\n", cnt - 1);
    }
    return 0;
}

#include <cstdio>

#include <algorithm>

using namespace std;

int v[1010][1010];

bool visit[1010];

int n;

void dfs(int node) {

visit[node] = true;

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

if(visit[i] == false && v[node][i] == 1)

dfs(i);

}

int main() {

int m, k, a, b;

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

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

scanf("%d%d", &a, &b);

v[a][b] = v[b][a] = 1;

}

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

fill(visit, visit + 1010, false);

scanf("%d", &a);

int cnt = 0;

visit[a] = true;

for(int j = 1; j <= n; j++) {

if(visit[j] == false) {

dfs(j);

cnt++;

}

printf("%d\n", cnt - 1);

}

return 0;

}

如何去除chrome最常访问的网页

chrome -> 更多工具 -> 扩展程序

Snip20160823_70

扩展程序 -> 获取更多扩展程序

Snip20160823_71

搜索Stylish -> 添加至chrome

Snip20160823_72

添加样式：名称随便写，代码部分写：[id=”most-visited”]{display:none}

应用对象：与该正则表达式匹配的网址，后面的文本框里面写.*\/chrome\/newtab.*

如果还想搜索框也去除，就在代码部分加一句#f{display:none}

然后点击保存

然后就完全去除啦~~~~~

Snip20160823_74

1076. Forwards on Weibo (30)-PAT甲级真题（图的遍历bfs）

Weibo is known as the Chinese version of Twitter. One user on Weibo may have many followers, and may follow many other users as well. Hence a social network is formed with followers relations. When a user makes a post on Weibo, all his/her followers can view and forward his/her post, which can then be forwarded again by their followers. Now given a social network, you are supposed to calculate the maximum potential amount of forwards for any specific user, assuming that only L levels of indirect followers are counted.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (<=1000), the number of users; and L (<=6), the number of levels of indirect followers that are counted. Hence it is assumed that all the users are numbered from 1 to N. Then N lines follow, each in the format:

M[i] user_list[i]

where M[i] (<=100) is the total number of people that user[i] follows; and user_list[i] is a list of the M[i] users that are followed by user[i]. It is guaranteed that no one can follow oneself. All the numbers are separated by a space.

Then finally a positive K is given, followed by K UserID’s for query.

Output Specification:

For each UserID, you are supposed to print in one line the maximum potential amount of forwards this user can triger, assuming that everyone who can view the initial post will forward it once, and that only L levels of indirect followers are counted.

Sample Input:
7 3
3 2 3 4
0
2 5 6
2 3 1
2 3 4
1 4
1 5
2 2 6
Sample Output:
4
5

题目大意：给出每个用户关注的人的id，和转发最多的层数，求一个id发了条微博最多会有多少个人转发
分析：带层数的广度优先，因为一个用户只能转发一次，所以用inq判断当前结点是否入队过了，如果入队过了就不能重复入队（重复转发消息），inq 邻接表v 都可以使用int只存储id，queue的数据类型必须为node，同时保存它的id和layer层数，控制不超过L层～

#include <cstdio>
#include <queue>
#include <vector>
using namespace std;
int n, l, m, k;
struct node {
    int id, layer;
};
vector<vector<int>> v;
int bfs(node tnode) {
    bool inq[1010] = {false};
    queue<node> q;
    q.push(tnode);
    inq[tnode.id] = true;
    int cnt = 0;
    while(!q.empty()) {
        node top = q.front();
        q.pop();
        int topid = top.id;
        for(int i = 0; i < v[topid].size(); i++) {
            int nextid = v[topid][i];
            if(inq[nextid] == false && top.layer < l) {
                node next = {nextid, top.layer + 1};
                q.push(next);
                inq[next.id] = true;
                cnt++;
            }
        }
    }
    return cnt;
}
int main() {
    scanf("%d %d", &n, &l);
    v.resize(n + 1);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &m);
        for(int j = 0; j < m; j++) {
            int temp;
            scanf("%d", &temp);
            v[temp].push_back(i);
        }
    }
    scanf("%d", &k);
    int tid;
    for(int i = 0; i < k; i++) {
        scanf("%d", &tid);
        node tnode = {tid, 0};
        printf("%d\n", bfs(tnode));
    }
    return 0;
}

#include <cstdio>

#include <queue>

#include <vector>

using namespace std;

int n, l, m, k;

struct node {

int id, layer;

};

vector<vector<int>> v;

int bfs(node tnode) {

bool inq[1010] = {false};

queue<node> q;

q.push(tnode);

inq[tnode.id] = true;

int cnt = 0;

while(!q.empty()) {

node top = q.front();

q.pop();

int topid = top.id;

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

int nextid = v[topid][i];

if(inq[nextid] == false && top.layer < l) {

node next = {nextid, top.layer + 1};

q.push(next);

inq[next.id] = true;

cnt++;

}

return cnt;

}

int main() {

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

v.resize(n + 1);

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

scanf("%d", &m);

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

int temp;

scanf("%d", &temp);

v[temp].push_back(i);

}

scanf("%d", &k);

int tid;

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

scanf("%d", &tid);

node tnode = {tid, 0};

printf("%d\n", bfs(tnode));

}

return 0;

}

1034. Head of a Gang (30)-PAT甲级真题（连通分量、图的遍历dfs）

One way that the police finds the head of a gang is to check people’s phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of all the phone calls made between the two persons. A “Gang” is a cluster of more than 2 persons who are related to each other with total relation weight being greater than a given threshold K. In each gang, the one with maximum total weight is the head. Now given a list of phone calls, you are supposed to find the gangs and the heads.

Input Specification:

Each input file contains one test case. For each case, the first line contains two positive numbers N and K (both less than or equal to 1000), the number of phone calls and the weight threthold, respectively. Then N lines follow, each in the following format:

Name1 Name2 Time

where Name1 and Name2 are the names of people at the two ends of the call, and Time is the length of the call. A name is a string of three capital letters chosen from A-Z. A time length is a positive integer which is no more than 1000 minutes.

Output Specification:

For each test case, first print in a line the total number of gangs. Then for each gang, print in a line the name of the head and the total number of the members. It is guaranteed that the head is unique for each gang. The output must be sorted according to the alphabetical order of the names of the heads.

Sample Input 1:
8 59
AAA BBB 10
BBB AAA 20
AAA CCC 40
DDD EEE 5
EEE DDD 70
FFF GGG 30
GGG HHH 20
HHH FFF 10
Sample Output 1:
2
AAA 3
GGG 3
Sample Input 2:
8 70
AAA BBB 10
BBB AAA 20
AAA CCC 40
DDD EEE 5
EEE DDD 70
FFF GGG 30
GGG HHH 20
HHH FFF 10
Sample Output 2:
0

题目大意：给出1000条以内的通话记录A B和权值w，和阈值k，如果一个团伙人数超过2人并且通话总权值超过k，令团伙里面的自身权值的最大值为头目，输出所有满足条件的团伙的头目，和他们团伙里面的人数
分析：总的来说是一个判断一个图的连通分量的个数，用图的遍历解决，深度优先遍历。
1.因为给的是字母，要用两个map把它们转换成数字，从1开始排列命名所有不同的人的id，存储在两个map里面，一个字符串对应id，一个id对应字符串，方便查找，正好顺便统计了总共的人数idNumber。
2.建立两个数组，weight和G，分别存储每条边的权值和每个结点的权值，因为这两个题目中都要求得后判断。
3.用传递引用的方法深度优先dfs，这样传入的参数在dfs后还能保存想要求得的值
4.遍历过一条边之后就把这条边的权值设为0（ G[u][v] = G[v][u] = 0;）防止出现回路遍历死循环

#include <iostream>
#include <map>
using namespace std;
map<string, int> stringToInt;
map<int, string> intToString;
map<string, int> ans;
int idNumber = 1, k;
int stoifunc(string s) {
    if(stringToInt[s] == 0) {
        stringToInt[s] = idNumber;
        intToString[idNumber] = s;
        return idNumber++;
    } else {
        return stringToInt[s];
    }
}
int G[2010][2010], weight[2010];
bool vis[2010];
void dfs(int u, int &head, int &numMember, int &totalweight) {
    vis[u] = true;
    numMember++;
    if(weight[u] > weight[head])
        head = u;
    for(int v = 1; v < idNumber; v++) {
        if(G[u][v] > 0) {
            totalweight += G[u][v];
            G[u][v] = G[v][u] = 0;
            if(vis[v] == false)
                dfs(v, head, numMember, totalweight);
        }
    }
}
void dfsTrave() {
    for(int i = 1; i < idNumber; i++) {
        if(vis[i] == false) {
            int head = i, numMember = 0, totalweight = 0;
            dfs(i, head, numMember, totalweight);
            if(numMember > 2 && totalweight > k)
                ans[intToString[head]] = numMember;
        }
    }
}
int main() {
    int n, w;
    cin >> n >> k;
    string s1, s2;
    for(int i = 0; i < n; i++) {
        cin >> s1 >> s2 >> w;
        int id1 = stoifunc(s1);
        int id2 = stoifunc(s2);
        weight[id1] += w;
        weight[id2] += w;
        G[id1][id2] += w;
        G[id2][id1] += w;
    }
    dfsTrave();
    cout << ans.size() << endl;
    for(auto it = ans.begin(); it != ans.end(); it++)
        cout << it->first << " " << it->second << endl;
    return 0;
}

#include <iostream>

#include <map>

using namespace std;

map<string, int> stringToInt;

map<int, string> intToString;

map<string, int> ans;

int idNumber = 1, k;

int stoifunc(string s) {

if(stringToInt[s] == 0) {

stringToInt[s] = idNumber;

intToString[idNumber] = s;

return idNumber++;

} else {

return stringToInt[s];

}

int G[2010][2010], weight[2010];

bool vis[2010];

void dfs(int u, int &head, int &numMember, int &totalweight) {

vis[u] = true;

numMember++;

if(weight[u] > weight[head])

head = u;

for(int v = 1; v < idNumber; v++) {

if(G[u][v] > 0) {

totalweight += G[u][v];

G[u][v] = G[v][u] = 0;

if(vis[v] == false)

dfs(v, head, numMember, totalweight);

}

void dfsTrave() {

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

if(vis[i] == false) {

int head = i, numMember = 0, totalweight = 0;

dfs(i, head, numMember, totalweight);

if(numMember > 2 && totalweight > k)

ans[intToString[head]] = numMember;

}

int main() {

int n, w;

cin >> n >> k;

string s1, s2;

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

cin >> s1 >> s2 >> w;

int id1 = stoifunc(s1);

int id2 = stoifunc(s2);

weight[id1] += w;

weight[id2] += w;

G[id1][id2] += w;

G[id2][id1] += w;

}

dfsTrave();

cout << ans.size() << endl;

for(auto it = ans.begin(); it != ans.end(); it++)

cout << it->first << " " << it->second << endl;

return 0;

}

1024. Palindromic Number (25)-PAT甲级真题（大整数相加）

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

Given any positive integer N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

Input Specification:

Each input file contains one test case. Each case consists of two positive numbers N and K, where N (<= 10^10) is the initial numer and K (<= 100) is the maximum number of steps. The numbers are separated by a space.

Output Specification:

For each test case, output two numbers, one in each line. The first number is the paired palindromic number of N, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after K steps, just output the number obtained at the Kth step and K instead.

Sample Input 1:
67 3
Sample Output 1:
484
2
Sample Input 2:
69 3
Sample Output 2:
1353
3
题目大意：给定一个数字，和允许翻转后相加的次数cnt，求要多少次才能变成一个回文数字，输出那个回文数字和翻转相加了多少次，如果本身就是回文数字就输出0次，如果超过给定的次数cnt了，就输出那个不是回文的结果，并且输出给定的次数cnt
分析：1、会超出long int类型（会有两个点溢出错误），所以用字符串存储，大整数相加
2、可以通过对字符串翻转后比较来判断是否为回文串

#include <iostream>
#include <algorithm>
using namespace std;
string s;
void add(string t) {
    int len = s.length(), carry = 0;
    for(int i = 0; i < len; i++) {
        s[i] = s[i] + t[i] + carry - '0';
        carry = 0;
        if(s[i] > '9') {
            s[i] = s[i] - 10;
            carry = 1;
        }
    }
    if(carry) s += '1';
    reverse(s.begin(), s.end());
}
int main() {
    int cnt, i;
    cin >> s >> cnt;
    for(i = 0; i <= cnt; i++) {
        string t = s;
        reverse(t.begin(), t.end());
        if(s == t || i == cnt) break;
        add(t);
    }
    cout << s << endl << i;
    return 0;
}

#include <iostream>

#include <algorithm>

using namespace std;

string s;

void add(string t) {

int len = s.length(), carry = 0;

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

s[i] = s[i] + t[i] + carry - '0';

carry = 0;

if(s[i] > '9') {

s[i] = s[i] - 10;

carry = 1;

}

if(carry) s += '1';

reverse(s.begin(), s.end());

}

int main() {

int cnt, i;

cin >> s >> cnt;

for(i = 0; i <= cnt; i++) {

string t = s;

reverse(t.begin(), t.end());

if(s == t || i == cnt) break;

add(t);

}

cout << s << endl << i;

return 0;

}