BZOJ1601&&洛谷P1550 [USACO08OCT]打井Watering Hole

kruskal裸题

我们将在每个点打井费用看作是从0号点连向当前点的一条边，每个井之间连边，然后跑一遍最小生成树，ans就是最终答案

代码

//By AcerMo
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int M=500;
struct edge
{
	int fr,to,cost;
	bool friend operator < (edge a,edge b)
	{
		return a.cost>b.cost;
	}
}add;
priority_queue<edge>q; 
int n,fa[M],siz[M];
inline int find(int x)
{
	if (x!=fa[x]) return fa[x]=find(fa[x]);
	return x;
}
inline int read()
{
	int x=0;char ch=getchar();
	while (ch>'9'||ch<'0') ch=getchar();
	while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x;
}
inline void unionn(int a,int b)
{
	if (siz[a]<siz[b]) siz[b]+=siz[a],fa[a]=b;
	else siz[a]+=siz[b],fa[b]=a;
	return ;
}
int main()
{
	n=read();add.fr=0;int ans=0,cnt=0;
	for (int i=0;i<=n;i++) fa[i]=i,siz[i]=1;
	for (int i=1;i<=n;i++)
		add.to=i,add.cost=read(),q.push(add);
	for (int i=1;i<=n;i++)
	for (int k=1;k<=n;k++)
	{
		add.fr=i;add.to=k;add.cost=read();
		if (i==k) continue;q.push(add);
	}
	while (q.size()&&cnt<n)
	{
		add=q.top();q.pop();
		int r1=find(add.fr);
		int r2=find(add.to);
		if (r1!=r2)	
			unionn(r1,r2),ans+=add.cost,cnt++;
	} 
	cout<<ans;
	return 0;
}