1046. Shortest Distance (20)

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

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main(){
	int N;
	scanf("%d",&N);
	int a[100005];
	memset(a,0,sizeof(a));
	int i,j,k;
	int tmp;
	for(i=1;i<N+1;i++){
		scanf("%d",&tmp);
		a[i]=a[i-1]+tmp;
	}
	int m;
	scanf("%d",&m);
	while(m--){
		int index1,index2;
		scanf("%d%d",&index1,&index2);
		if(index1>index2){
			tmp=index1;
			index1=index2;
			index2=tmp;
		}
		int d1=a[index2-1]-a[index1-1];
		int d2=a[index1-1]+a[N]-a[index2-1];
		if(d1>d2){
			printf("%d\n",d2);
		}else {
			printf("%d\n",d1);
		}
	}
	return 0;
}

  



posted @ 2017-10-29 09:21  gaoren  阅读(149)  评论(0编辑  收藏  举报