PAT-1046.Shortest Distance (20)

题目描述：

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

1 3

2 5

4 1

Sample Output:

分析：题目描述的场景可以形成一个环，考虑在一个环上点A到点B可以有两个不同方向，取这两个方向较小者。一开始想用Floyd算法，但貌似没有必要。

代码：http://blog.csdn.net/sunbaigui/article/details/8657186

#include<iostream>  
#include<cmath>  
using namespace std;  
#define MAX 100000  
int dis[MAX];  
  
int min(int a,int b){  
    if(a>b) return b;  
    else  
        return a;  
}  
  
int main()  
{  
    int N,M;  
    int total = 0; //如果变量没有初始化便进行运算，会出错。  
    int i,j,temp;  
    int begin,end;  
    cin>>N;  
    dis[1] = 0;  
    for(i=1; i<=N; i++){  
        cin>>temp;  
        total += temp;  
        dis[i+1] = dis[i] + temp;  
    }  
  
    cin>>M;  
    while(M--){  
        cin>>begin>>end;  
  
        temp = abs(dis[end] - dis[begin]);  
        cout<<min( temp,total-temp )<<endl;       
    }  
  
    return 0;  
}

补充：软件开发 , C++ ,