P1550 [USACO08OCT]打井Watering Hole

传送门

如果没有挖井只是联通所有点就是最小生成树

但是有挖井，我们需要转换一下

把挖井操作看成 0 号点连一条边过去

然后还是最小生成树..

因为是稠密图，所以用 prim 算法来求最小生成树

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
using namespace std;
typedef long long ll;
inline int read()
{int x=0,f=1; char ch=getchar();while(ch<'0'||ch>'9') { if(ch=='-') f=-1; ch=getchar(); }while(ch>='0'&&ch<='9') { x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); }return x*f;
}
const int N=507,INF=0x3f3f3f3f;
int n,ans,mp[N][N];
struct data
{int dis,id;inline bool operator < (const data &tmp) const {return dis>tmp.dis;}
};
priority_queue <data> q;
int dis[N];
bool vis[N];
void prim(int st)//prim模板
{memset(dis,INF,sizeof(dis));q.push((data){0,st}); dis[0]=0;while(!q.empty()){data x=q.top(); q.pop();if(vis[x.id]) continue;vis[x.id]=1; ans+=x.dis;for(int i=0;i<=n;i++)if(dis[i]>mp[x.id][i]){dis[i]=mp[x.id][i];q.push((data){mp[x.id][i],i});}}
}
int main()
{n=read();for(int i=1;i<=n;i++) mp[0][i]=mp[i][0]=read();//0号点连边for(int i=1;i<=n;i++)for(int j=1;j<=n;j++) mp[i][j]=mp[j][i]=read();prim(0);printf("%d",ans);return 0;
}

转载于:https://www.cnblogs.com/LLTYYC/p/9854655.html

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