题目地址：HDU 3639
先用强连通分量缩点，缩点之后，再重新按缩点之后的块逆序构图，，每个块的值是里边缩的点的个数，那么得到选票的最大的一定是重新构图后入度为0的块，然后求出来找最大值即可。
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 long long#define pi acos(-1.0)const int mod=1e9+7;const int INF=0x3f3f3f3f;const double eqs=1e-9;const int MAXN=5000+10;const int MAXM=30000+10;int head[MAXN], cnt, low[MAXN], dfn[MAXN], belong[MAXN], instack[MAXN], stk[MAXN], dp[MAXN];int ans, indx, top;struct node{int u, v, next;}edge[MAXM];void add(int u, int v){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–];belong[v]=ans;dp[ans]++;instack[v]=0;if(u==v) break;}}}void init(){memset(head,-1,sizeof(head));memset(dfn,0,sizeof(dfn));memset(instack,0,sizeof(instack));memset(dp,0,sizeof(dp));cnt=ans=indx=top=0;}int head1[MAXN], cnt1, vis[MAXN], in[MAXN], max1, sum;struct node1{int u, v, next;}edge1[MAXM];void add1(int u, int v){edge1[cnt1].v=v;edge1[cnt1].next=head1[u];head1[u]=cnt1++;}void dfs(int u){vis[u]=1;sum+=dp[u];for(int i=head1[u];i!=-1;i=edge1[i].next){int v=edge1[i].v;if(!vis[v]){dfs(v);}}}void init1(){memset(head1,-1,sizeof(head1));memset(vis,0,sizeof(vis));memset(in,0,sizeof(in));cnt1=0;max1=0;}int c[MAXN], tot, d[MAXN];int main(){int t, n, m, i, j, u, v, Case=0;scanf("%d",&t);while(t–){Case++;scanf("%d%d",&n,&m);init();while(m–){scanf("%d%d",&u,&v);add(u,v);}for(i=0;i<n;i++){if(!dfn[i]) tarjan(i);}init1();for(i=0;i<n;i++){for(j=head[i];j!=-1;j=edge[j].next){if(belong[i]!=belong[edge[j].v]){add1(belong[edge[j].v],belong[i]);in[belong[i]]++;}}}memset(d,-1,sizeof(d));for(i=1;i<=ans;i++){if(!in[i]){sum=0;memset(vis,0,sizeof(vis));dfs(i);d[i]=sum;max1=max(max1,d[i]);}}tot=0;for(i=0;i<n;i++){if(d[belong[i]]==max1)c[tot++]=i;}printf("Case %d: %d\n",Case, max1-1);for(i=0;i<tot;i++){printf("%d",c[i]);if(i!=tot-1) printf(" ");}puts("");}return 0;}
旅行是一种病。一旦感染了，你就再也无法摆脱。
