题目

链接

http://poj.org/problem?id=2421

字面描述

Constructing Roads
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 36942 Accepted: 16631
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
Source

PKU Monthly,kicc

重点处理

对于已经加好的路径边权赋值为0

代码实现

#include
#include
#include
#include
#include
using namespace std;

const int maxn=100+10;
const int inf=2e9;
int n,ans,m;
int lowcost[maxn];
int c[maxn][maxn];
bool vis[maxn];
inline void Prim(){
	//初始化 
	memset(vis,false,sizeof(vis));
	ans=0;
	vis[1]=true;
	for(int i=1;i<=n;i++)lowcost[i]=c[1][i];
	//n-1条边的添加 
	for(int i=1;i<n;i++){
		int temp=inf;
		int t=1;
		//寻找离已添加最近的节点t 
		for(int j=1;j<=n;j++){
			if(!vis[j]&&lowcost[j]<temp){
				temp=lowcost[j];
				t=j; 
			}
		}
		ans+=temp;
		if(t==1)break;
		vis[t]=true;
		//更新未添加节点 
		for(int j=1;j<=n;j++){
			if(!vis[j]&&c[t][j]<lowcost[j])lowcost[j]=c[t][j];
		}
	}
} 
int main(){
	scanf("%d",&n);
	memset(c,0x3f,sizeof(c));
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++)scanf("%d",&c[i][j]); 
	}
	scanf("%d",&m);
	for(int i=1;i<=m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		c[u][v]=0;
		c[v][u]=0;
	}
	Prim();
	printf("%d\n",ans);
	return 0;
} 
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