HDU 1827

Tarjan缩点后求入度为0的点，即是要求最少电话。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int cst[1010];
struct Edge{
	int u,v;
	int next;
}edge[2010];
int head[1010],dfn[1010],low[1010],stack[1010],st;
bool instack[1010];
int tot,nTime,ans,btn;
int n,m;
int belong[1010],label[1010];
int edsave[2010][2];
int degree[1010];
void addedge(int u,int v){
	edge[tot].u=u;
	edge[tot].v=v;
	edge[tot].next=head[u];
	head[u]=tot++;
}

void tarjan(int u,int ufa){
	dfn[u]=low[u]=++nTime;
	stack[++st]=u;
	instack[u]=true;
	for(int e=head[u];e!=-1;e=edge[e].next){
		int v=edge[e].v;
		if(dfn[v]==-1){
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
		}
		else if(instack[v]){
			low[u]=min(low[u],dfn[v]);
		}
	}
	if(dfn[u]==low[u]){
		int v; int mincst=(1<<30)-1;
		btn++;
		do{
			v=stack[st--];
			instack[v]=false;
			belong[v]=btn;
			mincst=min(mincst,cst[v]);
		}while(u!=v);
		label[btn]=mincst;
	}
}

int main(){
	int u,v;
	while(scanf("%d%d",&n,&m)!=EOF){
		memset(head,-1,sizeof(head));
		memset(dfn,-1,sizeof(dfn));
		memset(low,-1,sizeof(low));
		memset(instack,false,sizeof(instack));
		memset(degree,0,sizeof(degree));
		st=tot=nTime=ans=btn=0;
		for(int i=1;i<=n;i++)
		scanf("%d",&cst[i]);
		for(int i=1;i<=m;i++){
			scanf("%d%d",&u,&v);
			edsave[i][0]=u; edsave[i][1]=v;
			addedge(u,v);
		}
		for(int i=1;i<=n;i++){
			if(dfn[i]==-1){
				tarjan(i,0);
			}
		}
		for(int i=1;i<=m;i++){
			if(belong[edsave[i][0]]==belong[edsave[i][1]])
			continue;
			else degree[belong[edsave[i][1]]]++;
		}
		int cnt=0;
		for(int i=1;i<=btn;i++)
		if(degree[i]==0){
			cnt++;
			ans+=label[i];
		}
		printf("%d %d\n",cnt,ans);
	}
	return 0;
}