1046. Shortest Distance (20)-PAT甲级真题

The task is really simple: given N exits on a highway which forms a simple cycle, you are supposed to tell the shortest distance between any pair of exits.

Input Specification:

Each input file contains one test case. For each case, the first line contains an integer N (in [3, 105]), followed by N integer distances D1 D2 … DN, where Di is the distance between the i-th and the (i+1)-st exits, and DN is between the N-th and the 1st exits. All the numbers in a line are separated by a space. The second line gives a positive integer M (<=104), with M lines follow, each contains a pair of exit numbers, provided that the exits are numbered from 1 to N. It is guaranteed that the total round trip distance is no more than 107.

Output Specification:

For each test case, print your results in M lines, each contains the shortest distance between the corresponding given pair of exits.

Sample Input:
5 1 2 4 14 9
3
1 3
2 5
4 1
Sample Output:
3
10
7

分析：简单模拟。所有结点连起来会形成一个环形，dis[i]存储第1个结点到第i个结点的下一个结点的距离，sum保存整个路径一圈的总和值。求得结果就是dis[right – 1] – dis[left – 1]和 sum – dis[right – 1] – dis[left – 1]中较小的那一个～～
注意：可能left和right的顺序颠倒了，这时候要把left和right的值交换～

#include <iostream>
#include <vector>
using namespace std;
int main() {
    int n;
    scanf("%d", &n);
    vector<int> dis(n + 1);
    int sum = 0, left, right, cnt;
    for(int i = 1; i <= n; i++) {
        int temp;
        scanf("%d", &temp);
        sum += temp;
        dis[i] = sum;
    }
    scanf("%d", &cnt);
    for(int i = 0; i < cnt; i++) {
        scanf("%d %d", &left, &right);
        if(left > right)
            swap(left, right);
        int temp = dis[right - 1] - dis[left - 1];
        printf("%d\n", min(temp, sum - temp));
    }
    return 0;
}

#include <iostream>

#include <vector>

using namespace std;

int main() {

int n;

scanf("%d", &n);

vector<int> dis(n + 1);

int sum = 0, left, right, cnt;

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

int temp;

scanf("%d", &temp);

sum += temp;

dis[i] = sum;

}

scanf("%d", &cnt);

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

scanf("%d %d", &left, &right);

if(left > right)

swap(left, right);

int temp = dis[right - 1] - dis[left - 1];

printf("%d\n", min(temp, sum - temp));

}

return 0;

}

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