HDU 3639 Hawk-and-Chicken （强连通分量+树形DP）

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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;}

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