题目地址：HDU 1827
先缩点，，缩完点后，找出入度为0的块就是需要传递的块。然后用块中花费最少的来当代表块中的花费。累加起来就行了。
代码如下：
#include <iostream>#include <string.h>#include <math.h>#include <queue>#include <algorithm>#include <stdlib.h>#include <map>#include <set>#include <stdio.h>using namespace std;#define LL __int64#define pi acos(-1.0)const int mod=1e9+7;const int INF=0x3f3f3f3f;const double eqs=1e-9;const int MAXN=1000+10;const int MAXM=2000+10;int head[MAXN], cnt, indx, top, ans;int low[MAXN], dfn[MAXN], belong[MAXN], instack[MAXN], stk[MAXN], cost[MAXN], d[MAXN], in[MAXN];struct node{int u, v, next;}edge[MAXM];void add(int u, int v){edge[cnt].u=u;edge[cnt].v=v;edge[cnt].next=head[u];head[u]=cnt++;}void tarjan(int u){low[u]=dfn[u]=++indx;instack[u]=1;stk[++top]=u;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].v;if(!dfn[v]){tarjan(v);low[u]=min(low[u],low[v]);}else if(instack[v]){low[u]=min(low[u],dfn[v]);}}if(low[u]==dfn[u]){ans++;while(1){int v=stk[top–];instack[v]=0;belong[v]=ans;if(u==v) break;}}}void init(){memset(head,-1,sizeof(head));memset(instack,0,sizeof(instack));memset(dfn,0,sizeof(dfn));memset(cost,INF,sizeof(cost));memset(in,0,sizeof(in));cnt=top=indx=ans=0;}int main(){int n, m, i, u, v, sum, tot;while(scanf("%d%d",&n,&m)!=EOF){init();for(i=1;i<=n;i++){scanf("%d",&d[i]);}while(m–){scanf("%d%d",&u,&v);add(u,v);}for(i=1;i<=n;i++){if(!dfn[i]) tarjan(i);}//printf("ans=%d\n",ans);for(i=0;i<cnt;i++){if(belong[edge[i].u]!=belong[edge[i].v])in[belong[edge[i].v]]++;}sum=tot=0;for(i=1;i<=n;i++){if(cost[belong[i]]>d[i])cost[belong[i]]=d[i];}for(i=1;i<=ans;i++){if(!in[i]){tot++;sum+=cost[i];}}printf("%d %d\n",tot,sum);}return 0;}
学会宽容，要有一颗宽容的爱心！
