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[];
 int main(){
     //freopen("D:\\INPUT.txt","r",stdin);
     int n;
     scanf("%d",&n);
     int i;
     dis[]=;
     int total=;
     for(i=;i<=n;i++){
         scanf("%d",&dis[i]);
         dis[i]=dis[i]+dis[i-];
     }
     scanf("%d",&dis[]);
     total=dis[n]+dis[];//一个环总长度
     //cout<<total<<endl;
     int m;
     scanf("%d",&m);
     int a,b;
     for(i=;i<m;i++){
         scanf("%d %d",&a,&b);
         if(a>b){
             a=a+b;
             b=a-b;
             a=a-b;
         }
         printf("%d\n",dis[b]-dis[a]<total-(dis[b]-dis[a])?dis[b]-dis[a]:total-(dis[b]-dis[a]));
     }
     return ;
 }