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

分析

这题在读入距离数据时把它按照从1到它的距离储存起来，后面求最小距离时就是min(总距离-dist[i][j],dist[i][j])

#include<iostream>
#include<math.h>
using namespace std;
int dist[100001]={0};
int main(){
	int n,dis,s,e,m;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>dis;
		dist[i+1]=dist[i]+dis;
	}
	cin>>m;
	for(int i=0;i<m;i++){
		cin>>s>>e;
		int t1,t2;
		t1=min(s,e);
		t2=max(s,e);
		cout<<min(dist[t2]-dist[t1],dist[n+1]-(dist[t2]-dist[t1]))<<endl;
	}
	return 0;
}