PAT (Advanced Level) Practice 1046 Shortest Distance（20分）【查找】

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 $[3,10^5]$), followed by $N$ integer distances $D_1 D_2 ⋯ D_N, where D_i$ is the distance between the $i$-th and the $(i+1)$-st exits, and $D_N$ 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 $M (≤10^4)$, 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 $10^7$.

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

题意

给出一个环，环上有一些点。给出相邻两个点之间的距离。查询任意两点之间的最小距离。

思路

对数据进行预处理，在输入时记录每个点到第1个点的距离dist[i]，由于是一个圆环，所以两点之间只有两条路径，判断一下即可。

代码

#include <iostream>
#include <cstdio>

using namespace std;

int a[100005], dis[100005];

int main() {
    dis[1] = 0;

    int n, sum = 0;
    scanf("%d", &n);

    for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
        sum += a[i];
        dis[i + 1] = dis[i] + a[i];
    }

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

    while (m--) {
        int b1, b2;
        scanf("%d%d", &b1, &b2);
        if (b2 < b1) swap(b1, b2);

        int s1 = dis[b2] - dis[b1];
        int s2 = sum - s1;
        printf("%d", s1 < s2 ? s1 : s2);

        if (m) putchar('\n');
    }
}