1046 Shortest Distance (20 分)（模拟）

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：

简单模拟。所有结点连起来会形成一个环形。dis[i]存储第1个结点到第i个结点的下一个结点的距离。sum保存整个路径一圈的总和值。求得结果就是dis[right – 1] – dis[left – 1]和 sum – dis[right – 1] – dis[left – 1]中较小的那一个～～

注意：

可能left和right的顺序颠倒了，这时候要把left和right的值交换～

原文链接：https://blog.csdn.net/liuchuo/article/details/52108507

超时题解

额，就不知为何超时，，，

#include <bits/stdc++.h>

using namespace std;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,m,sum=0,a,b;
    scanf("%d",&n);
    vector<int> v(n+1);
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&v[i]);
        sum+=v[i];
    }
    scanf("%d",&m);
    for(int i=1; i<=m; i++)
    {
        int temp=0;
        scanf("%d%d",&a,&b);
        if(a>b) swap(a,b);
        for(int j=a;j<b;j++){
            temp+=v[j];
        }
        printf("%d\n",temp<sum-temp?temp:sum-temp);
    }
    return 0;
}

正确题解

正好刚做完一个用累加和然后相减的方法求两个坐标之间差值的题，这里用的方法也一样啦~
求得结果就是dis[right – 1] – dis[left – 1]和 sum – dis[right – 1] – dis[left – 1]中较小的那一个～

#include <bits/stdc++.h>

using namespace std;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,m,sum=0,a,b;
    scanf("%d",&n);
    vector<int> v(n+1);
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&v[i]);
        sum+=v[i];
        //这个地方的小技巧就是累加
        v[i]+=v[i-1];
    }
    scanf("%d",&m);
    for(int i=1; i<=m; i++)
    {
        scanf("%d%d",&a,&b);
        if(a>b) swap(a,b);
        int temp=v[b-1]-v[a-1];
        printf("%d\n",temp<sum-temp?temp:sum-temp);
    }
    return 0;
}