题意：http://acm.hdu.edu.cn/showproblem.php?pid=1827

Summer Holiday

12 16
2 2 2 2 2 2 2 2 2 2 2 2 
1 3
3 2
2 1
3 4
2 4
3 5
5 4
4 6
6 4
7 4
7 12
7 8
8 7
8 9
10 9
11 10

3 6

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#include<vector>
int min(int a,int b)
{
	return a>b?b:a;
}
#define N 1010 
//N为最大点数  
#define M 2010 
//M为最大边数  
int n, m;//n m 为点数和边数  
struct Edge{  
    int from, to, nex;  
    bool sign;//是否为桥  
}edge[M<<1];  
int head[N], edgenum;  
void add(int u, int v){//边的起点和终点  
    Edge E={u, v, head[u], false};  
    edge[edgenum] = E;  
    head[u] = edgenum++;  
}  

int DFN[N], Low[N], Stack[N], top, Time; //Low[u]是点集{u点及以u点为根的子树} 中(所有反向弧)能指向的(离根最近的祖先v) 的DFN[v]值(即v点时间戳)  
int taj;//连通分支标号，从1开始  
int Belong[N];//Belong[i] 表示i点属于的连通分支  
bool Instack[N];  
vector<int> bcc[N]; //标号从1开始  

void tarjan(int u ,int fa)
{    
    DFN[u] = Low[u] = ++ Time ;    
    Stack[top ++ ] = u ;    
    Instack[u] = 1 ;   
    for (int i = head[u] ; ~i ; i = edge[i].nex )
	{    
        int v = edge[i].to;    
        if(DFN[v] == -1)  
        {    
            tarjan(v , u);    
            Low[u] = min(Low[u] ,Low[v]) ;  
            if(DFN[u] < Low[v])  
            {  
                edge[i].sign = 1;//为割桥  
            }  
        }    
        else if(Instack[v])
		{
			Low[u] = min(Low[u] ,DFN[v]) ;        
		}
    }    
    if(Low[u] == DFN[u])
	{    
        int now;  
        taj ++ ; 
		bcc[taj].clear();  
        do{  
            now = Stack[-- top] ;    
            Instack[now] = 0 ;   
            Belong [now] = taj ;  
            bcc[taj].push_back(now);  
        }while(now != u) ;  
    }  
}  

void tarjan_init(int all){  
    memset(DFN, -1, sizeof(DFN));  
    memset(Instack, 0, sizeof(Instack));  
    top = Time = taj = 0;  
    for(int i=1;i<=all;i++)
		if(DFN[i]==-1 )
			tarjan(i, i); //注意开始点标！！！  
}  
vector<int>G[N];  
int du[N];   //入度
void suodian(){  
    memset(du, 0, sizeof(du));  
    for(int i = 1; i <= taj; i++)
		G[i].clear();  
    for(int i = 0; i < edgenum; i++)
	{  
        int u = Belong[edge[i].from], v = Belong[edge[i].to];  
        if(u!=v)
		{
			G[u].push_back(v), du[v]++;  
		}
    }  
}  

void init(){memset(head, -1, sizeof(head)); edgenum=0;}  
 

int spd[1010];
int main()
{ 
	while(~scanf("%d%d",&n,&m))
	{
		init();
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&spd[i]); 
		}

		for(int i=0;i<m;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v); 
			add(u,v);
		}
		tarjan_init(n);
		suodian();
		int numb=0;
		int ans=0; 

		for(int i=1;i<=taj;i++)
		{
			if(du[i]==0&&bcc[i].size())
			{
				int minn=-19999;
				for(int j=0;j<bcc[i].size();j++)
				{
					if(spd[bcc[i][j]]<minn||minn==-19999)
					{
						minn=spd[bcc[i][j]];
					}
				}
				ans+=minn;
				 
				numb++;
			}
		}
		printf("%d %d\n",numb,ans);
	}
	return 0;
}
 

hdu 1827 Summer Holiday tarjan+缩点

原文：http://blog.csdn.net/u013532224/article/details/46832267