最小生成树 prime poj1258

题意:给你一个矩阵M[i][j]表示i到j的距离 求最小生成树

思路:裸最小生成树 prime就可以了

最小生成树专题

AC代码:

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
const int N=50000;
struct Edge{
    int fr,to;
    int w;
    friend bool operator< (Edge a, Edge b){
        return a.w>b.w;
    }
};
vector<Edge>Map[N];

void Prime(int n){
    int ans=0;
    Edge now;
    bool vis[N];
    mem(vis);
    priority_queue<Edge> Q;
    while(!Q.empty()) Q.pop();
    for(int i=0; i<Map[1].size(); ++i) Q.push(Map[1][i]);
    vis[1]=1;
    n--;
    while(n--){
        now=Q.top();
        Q.pop();
        if(vis[now.to])while(vis[now.to]){
            now=Q.top();
            Q.pop();
        }
        ans+=now.w;
        vis[now.to]=1;
        for(int i=0; i<Map[now.to].size(); ++i)
            if(!vis[Map[now.to][i].to]) Q.push(Map[now.to][i]);
    }
    printf("%d\n",ans);
}

void Add(int u,int v,int w){
    Edge e;
    e.fr=u,e.to=v,e.w=w;
    Map[u].push_back(e);
}
int main(){
    int t,n,u,v,w;
    while(~scanf("%d",&n)){
        mem(Map),mem(p);
        for(u=1; u<=n; ++u){
            for(v=1; v<=n; ++v){
            scanf("%d",&w);
            if(u!=v)
            Add(u,v,w),Add(v,u,w);
            }
        }
        Prime(n);
    }
    return 0;
}

 

posted on 2016-11-18 19:01  lazzzy  阅读(380)  评论(0编辑  收藏  举报

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