3.1简单模拟A1046. 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 <stdio.h>#include <algorithm>using namespace std;/*    本题有难度，得经常看看，很容易忘记*/int main(){//    int N=5;    int N;//输入的个数；    scanf("%d",&N);//    int dis[N+1]={0},a[N+1]={0,1,2,4,14,9};//之所以用N+1，是计算a[1]的时候需要用a[0]来补充    int dis[N+1]={0},a[N+1]={0};//dis[N+1]存放的是第1个结点到第i个结点的下一个结点的距离，之所以下一个，考虑循环问题；a[N+1]表示相邻两结点的距离,a[1]表示结点1和结点2之间的距离    for(int i=1;i<=N;++i){        scanf("%d",&a[i]);    }    int sum=0;    /*        一直报运行超时，参考答案用了很妙的方法，直接用动态的sum来算dis,这样就不用二重循环；    for(int i=1;i<=N;++i){        sum+=a[i];//用来后期比较顺时针和逆时针的和大小        for(int j=1;j<=i;++j){            dis[i]+=a[j];        }    }    */    for(int i=1;i<=N;++i){        sum +=a[i];        dis[i]=sum;    }//    int M=3;    int M;//需要比较的次数；    scanf("%d",&M);    int l,m;//存放输入的两个结点坐标；    int p,q;//分别存放两点之间的顺时针距离和逆时针距离；    int shortest=0;    for(int i=1;i<=M;++i){        scanf("%d%d",&l,&m);        if(l>m)swap(l,m);        p = dis[m-1]-dis[l-1];        q=sum-p;        shortest = min(p,q);        printf("%d\n",shortest);    }    return 0;}