1046. Shortest Distance (20)

迈不过友情╰ 2022-05-31 12:54 127阅读 0赞

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

--------------------

题目大意：

这个任务非常简单：给定一个高速公路上的N个出口，它形成一个简单的循环，你应该求出两个出口之间最短的距离。  
输入规格：  
每个文件包含一个测试用例。对于每种情况，第一行包含一个正整数N(在\[3,10的5次方\])，然后是N个整数距离D1 D2 ... DN，Di是第i个和第（i+1）个出口之间的距离，Dn位于第n个出口和第1个出口之间。直线上得所有数字用空格隔开。第二行给出一个正整数M（<=10的4次方），接下来是M行，每行都包含一对出口编号，给定的出口从1到N。保证一圈的距离不超过10的7次方。  
输出规范：  
对于每个测试用例，将结果打印在M行中，每一行都包含相应的给定出口之间的最短距离。

代码：

#include<stdio.h>
    int arr[100001];
    int main()
    {
        int i,j,n,m,k,t,cnt,x,y;
        scanf("%d",&n);
        cnt=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&m);
            cnt+=m;
            arr[i+1]=cnt;
        }
        scanf("%d",&k);
        for(i=1;i<=k;i++)
        {
            scanf("%d %d",&x,&y);
            if(x>y)
            {
                t=x;
                x=y;
                y=t;
            }
            if((cnt-arr[y]+arr[x])>(arr[y]-arr[x]))
            {
               printf("%d\n",arr[y]-arr[x]);
            }
            else
            {
                printf("%d\n",cnt-arr[y]+arr[x]);
            }
        }
        return 0;
    }