POJ

题意：给出一张图然后求最小生成树
输入：N个点(输入以0结尾)，然后对应N个点相连的 边数m, 以及对应指向的结点 以及权值
思路：采用prim算法来做，先类似于dijsktra，搜索集合外点的路径，最小的路径然后加更新集合。再采用优先队列优化，减小每次判断要路径权值大小

完整代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <string>
#include <queue>
using namespace std;
typedef pair<int,int> pii;
const int maxm = 1e5;
const int inf = 0x3f3f3f3f;
const int maxn = 30;
int top;
int n,m;
int ans;
int head[maxn];
int vis[maxn];
int dist[maxn];
//堆重载 
struct cmp{
    bool operator() (pii a, pii b){
        return a.first>b.first;
    }
};
struct Edge{
    int u,v,w;
    int next;
}edge[maxm];

void init(){
    memset(head,-1,sizeof(head));
    memset(edge,0,sizeof(edge));
    memset(dist,-1,sizeof(dist));
    memset(vis,0,sizeof(vis));
    top = 0;
    ans = 0;
}
void add(int v,int u,int w){
    edge[top].u = u;
    edge[top].v = v;
    edge[top].w = w;
    edge[top].next = head[u];
    head[u] = top++;
}
void prim(int s){
    int i;
    priority_queue<pii,vector<pii> ,cmp>Q;
    dist[s] = 0;
    vis[s] = 1;
    for(i = head[s]; ~i; i = edge[i].next){
        //先把到起点边的距离初始化为其本身 
        dist[edge[i].v] = edge[i].w;
        Q.push(make_pair(dist[edge[i].v],edge[i].v));
    }
    while(!Q.empty()){
        pii t = Q.top();
        Q.pop();
        if(vis[t.second]) continue;
        ans += t.first;
        vis[t.second] = 1;
        for(i = head[t.second]; ~i ;i = edge[i].next){
            int j = edge[i].v;
            //没被访问没有连接或者权值太小 
            if(!vis[j]&&(dist[j]>edge[i].w||dist[j]==-1)){
                dist[j] = edge[i].w;
                Q.push(make_pair(dist[j],j));
            }
        }
    }
}
int main(){
    while(cin>>n&&n){
        init();
        int w;
        char ch,op;
        for(int i=0;i<n-1;i++){        
            cin>>ch;
            cin>>m;
            for(int j=0;j<m;j++){
                cin>>op>>w;
                add(ch-'A',op-'A',w);
                add(op-'A',ch-'A',w);
            }    
        }
        prim(0);    
        cout<<ans<<endl;
    }
}