Constructing Roads——最小生成树Prim

Problem Description

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input

3 0 990 692 990 0 179 692 179 0 1 1 2

Sample Output

179

思路

被hdu的输入伤害了无数次，题目没有说清楚数据大小，也没有说是单样例输入还是样例。按照题目要求输入的我wa哭好多次，去看别人的博客才知道这输入简直。。。

最小生成树，与dijkstra算法的原理很像，但是更新那一步的扩展根dijkstra的就不同了，一不小心了很容易写错。

dijkstra算法的是dis[j]=min(dis[j],dis[index]+mp[index][j]);

Prim算法的是dis[j]=min(dis[j],mp[index][j]);   //是不用加上原来的路径长度的。

代码

#include<bits/stdc++.h>
using namespace std;
int mp[1005][1005];
int dis[1005],vis[1005];

int main(){
	int n,m,a,b,sum=0;
	while(cin>>n){
		sum=0;
		memset(vis,0,sizeof(vis));
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				cin>>mp[i][j];
			}
		}
		cin>>m;
		for(int i=0;i<m;i++){
			cin>>a>>b;
			mp[a][b]=mp[b][a]=0;
		}
		
		for(int i=1;i<=n;i++){
			dis[i]=mp[1][i];
		}
		vis[1]=1;
		
		for(int i=1;i<n;i++){
			int minn=2e9,index=-1;
			for(int j=1;j<=n;j++){
				if(minn>dis[j]&&!vis[j]){
					minn=dis[j];
					index=j;
				}
			}
			sum+=minn;
			vis[index]=1;
			for(int j=1;j<=n;j++){
				if(!vis[j]&&dis[j]>mp[index][j]){
					dis[j]=mp[index][j];
				}
			}
		}
		cout << sum <<endl;
	}
	
	return 0;
}