简单的全源最短路径

问题

1046 Shortest Distance (20 分)
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 [3,10
5
]), followed by N integer distances D
1
​
D
2
​
⋯ 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 (≤10
4
), 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 10
7
.

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
结尾无空行
Sample Output:
3
10
7
结尾无空行

思路

一开始的想法是用floyd算法，但是发现题目所给结点数远远超过200，肯定会超时。于是采用暴力解法，发现还是会超时。原因在于数据没有预处理。

代码

#include<bits/stdc++.h>
using namespace std;

const int maxn = 100010;

int dis[maxn];

int n,m;

int main(){
    cin>>n;
    int temp,total = 0;
    for (int i = 1; i <= n; ++i) {
        cin>>temp;
        total = total + temp;
        dis[i] = total;
    }
    cin>>m;
    int c1,c2;
    for (int j = 0; j < m; ++j) {
        cin>>c1>>c2;
        int ans1,ans2;
        if(c1<c2){
            ans1 = dis[c2-1] - dis[c1-1];
            ans2 = total - ans1;
        }
        else{
            ans1 = dis[c1-1] - dis[c2-1];
            ans2 = total - ans1;
        }
        if(ans1<ans2)
            cout<<ans1<<endl;
        else
            cout<<ans2<<endl;
    }
    return 0;
}

总结

在输入数据的时候就计算每一步的边权之和，并且存储在数组dis[]种。