PAT--1046. Shortest Distance

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

题解

容易想到前缀和，正向（顺时针）的容易计算，但反向(逆时针)的怎么算呢？
把环展开成一条线，即在a[1...n]后面重复一遍a[]，对于p, q两点，最小距离即为min(s[q - 1] - s[p - 1], s[p + n - 1] - s[q - 1])。

#include <bits/stdc++.h>using namespace std;const int maxn = 100010;int a[2 * maxn], s[2 * maxn];int n, m;int main(){#ifndef ONLINE_JUDGEfreopen("data.in", "r", stdin);#endif // ONLINE_JUDGE    scanf("%d", &n);    for(int i = 1; i <= n; ++i){        scanf("%d", a + i);        s[i] = s[i - 1] + a[i];    }    for(int i = n + 1; i <= 2 * n; ++i){        a[i] = a[i - n];        s[i] = s[i - 1] + a[i];    }    scanf("%d", &m);    int p, q;    while(m--){        scanf("%d %d", &p, &q);        if(p > q) swap(p, q);        int v1 = s[q - 1] - s[p - 1];        int v2 = s[p + n - 1] - s[q - 1];        printf("%d\n", min(v1, v2));    }    return 0;}