hdu1102之prim(堆优化)解法

分析:根据题意,显然利用最小生成树解法比较方便,直接利用prim解法写起来方便一些,利用优先队列进行堆优化,可以减小复杂度.

ps:

这题好坑啊,说的是一组数据,结果是多组数据.............

先上模板:

//int V;
//int ans;
//int mincost[MAXN];
//int used[MAXN];

//struct edge{ 
//int to,cost;
//edge (){}
//edge (int to,int cost):to(to),cost(cost){}
//};
//vector <edge> G[MAXN];//存放从i出发的边

//typedef pair<int,int> P;//first存放距离,second存放节点
//priority_queue<P,vector<P>,greater<P>> que;

void prim()
{
	//初始化
    memset(used,0,sizeof(used));
    fill(mincost,mincost+V,INF);
    ans = 0;
    que.push(P(0,0));
    
    while (!que.empty()){
    	//取最短边
        P p = que.top();
        que.pop();
        int d = p.first,v = p.second;
        if (used[v]) continue;
        used[v] = 1;
        ans += d;
        //以最短边的节点为基础,更新最短距离
        for (int i = 0; i < G[v].size(); i += 1){  
        	edge e = G [v][i];    
            if ((mincost[e.to] > e.cost) && !used[i]){
            	mincost[e.to] = e.cost;
            	que.push(P(mincost[e.to],e.to));
            }
        }
    }
    printf("%d\n",ans);
}	

直接上代码,练习下模板使用

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <stack>
#include <queue>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <bitset>
#include <list>
#include <sstream>
#include <set>
#include <functional>
using namespace std;

#define INF 0x3f3f3f3f
#define MAXN 120
typedef long long ll;
int n,q;
int V;
int res;
int mincost[MAXN];
int used[MAXN];

struct edge{ 
int to,cost;
edge (){}
edge (int to,int cost):to(to),cost(cost){}
};
vector <edge> G[MAXN];//存放从i出发的边

typedef pair<int,int> P;//first存放距离,second存放节点
priority_queue<P,vector<P>,greater<P>> que;

void prim()
{
    memset(used,0,sizeof(used));
    fill(mincost,mincost+V,INF);
    res = 0;
    que.push(P(0,0));
    
    while (!que.empty()){
        P p = que.top();
        que.pop();
        int v = p.second,d = p.first;
        if (used[v]) continue;
        used[v] = 1;
        res += d;
        
        for (int i = 0; i < G[v].size(); i += 1){  
        	edge e = G [v][i];    
            if ((mincost[e.to] > e.cost) && !used[i]){
            	mincost[e.to] = e.cost;
            	que.push(P(mincost[e.to],e.to));
            }
        }
    }
    printf("%d\n",res);
}

int main()
{
	while (scanf("%d",&n) != EOF){
		memset(G,0,sizeof(G));//每次要初始化啊
		for (int i = 0; i < n; i += 1){
			for (int j = 0; j < n; j += 1){
				int a;
				cin >> a;
				G[i].push_back(edge(j,a));
			}
		}
		cin >> q;
		for (int i = 0; i < q; i += 1){
			int a,b;
			cin >> a >> b;
			G[a-1][b-1].cost = G[b-1][a-1].cost = 0;
		}
		V = n;
		prim();
	}
	return 0;
}