hdu 5934

题解：建有向图，trajan缩点求强连通分量，板子题

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std; 
#define ll long long
#define MAX 10001 

int n,m;

//前向星
struct Edge
{
	int to,next;
}edge[1005*1005];
int head[MAX*2],tol;
void add(int a,int b)
{
	edge[tol].to = b;
	edge[tol].next = head[a];
	head[a] = tol ++;
}

//tarjan算法
int dfn[MAX*2],low[MAX*2],stack[MAX*2],belong[MAX*2],time,top,cnt;
bool instack[MAX*2];
int in[MAX*2],out[MAX*2];//记录缩点后的入度出度
void tarjan(int u)
{
	int v;
	dfn[u] = low[u] = ++time;
	stack[top ++] = u;
	instack[u] = true;
	for(int j = head[u] ; j != -1; j = edge[j].next)
	{
		v = edge[j].to;
		if(!dfn[v])
		{
			tarjan(v);
			if(low[v] < low[u])
				low[u] = low[v];
		}
		else if(instack[v] && dfn[v] <low[u])
			low[u] = dfn[v];
	}
	if(dfn[u] == low[u])
	{
		cnt ++;
		do
		{
			v = stack[-- top];
			instack[v] = false;
			belong[v] = cnt;
		}while(v != u);
	}
}
ll x[1005],y[1005],r[1005];
int val[1005];
int main()
{
	int a,b;
	int T;
	scanf("%d",&T);
	int Case=0;
	while(T --)
	{
		Case++;
		scanf("%d",&n);
		for(int i=1;i<=n;i++)
		{
			scanf("%lld%lld%lld%d",&x[i],&y[i],&r[i],&val[i]);
		}
		//建图
		tol = 0;
		memset(head,-1,sizeof(head));
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n;j++)
			{
				if(i==j)continue;
				ll dist=(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]);
				if(dist<=r[i]*r[i])
				add(i,j);
			}
		}
		//缩点
		cnt = time = top = 0;
		memset(dfn,0,sizeof(dfn));
		for(int i = 1; i <= n; i ++)
			if(!dfn[i])
				tarjan(i);
		memset(in,0,sizeof(in));
		memset(out,0,sizeof(out));
		for(int i = 1; i <= n; i ++)
		{
			for(int k = head[i]; k != -1; k = edge[k].next)
			{
				int j = edge[k].to;
				if(belong[i] != belong[j])
					out[belong[i]] ++, in[belong[j]] ++;
			}
		}
		
		//缩点之后找叶子和根的数量ll
		int ans=0;
		for(int i = 1; i <= cnt; i ++)
		{
			if(in[i]==0)
			{
				int mi=1e9;
				for(int j=1;j<=n;j++)
				{
					if(belong[j]==i)
					mi=min(mi,val[j]);
				}
				ans+=mi;
			}
		}
		 printf("Case #%d: %d\n",Case,ans);
	
	}
	return 0;
}