pat1046. Shortest Distance (20)

1046. Shortest Distance (20)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 ... 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 (<=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

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提交代码

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<cmath>
#include<string>
using namespace std;
int dis[100005];
int main(){
    //freopen("D:\INPUT.txt","r",stdin);
    int n;
    scanf("%d",&n);
    int i;
    dis[1]=0;
    int total=0;
    for(i=2;i<=n;i++){
        scanf("%d",&dis[i]);
        dis[i]=dis[i]+dis[i-1];
    }
    scanf("%d",&dis[0]);
    total=dis[n]+dis[0];//一个环总长度
    //cout<<total<<endl;
    int m;
    scanf("%d",&m);
    int a,b;
    for(i=0;i<m;i++){
        scanf("%d %d",&a,&b);
        if(a>b){
            a=a+b;
            b=a-b;
            a=a-b;
        }
        printf("%d
",dis[b]-dis[a]<total-(dis[b]-dis[a])?dis[b]-dis[a]:total-(dis[b]-dis[a]));
    }
    return 0;
}