HDU 4685 Prince and Princess（二分图 + 强连通）

2
4 4
2 1 2
2 1 2
2 2 3
2 3 4
1 2
2 1 2

Case #1:
2 1 2
2 1 2
1 3
1 4
Case #2:
2 1 2

感觉参考博客写的听仔细的

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 20005

int boy[N],girl[N],vis[N];  //   二分图
int time[N],low[N],cnt,time_num,instack[N],type[N]; //强连通
int n,m,mn;
vector<int>g[N];
stack<int>q;

bool dfs(int x)
{
    int i,j;

    fre(i,0,g[x].size())
    {
        int to=g[x][i];

        if(vis[to]) continue;

        vis[to]=1;

        if(girl[to]==0||dfs(girl[to]))
        {
            boy[x]=to;
            girl[to]=x;
            return true;
        }
    }
   return false;
}

void xiong()
{
    int i,j;
    mem(boy,0);
    mem(girl,0);

    fre(i,1,mn+1)
    {
        mem(vis,0);
        dfs(i);
    }
}


void leave()
{
    int i,j;
    int all=2*mn;

    fre(i,1,mn+1)
    {
        if(boy[i]==0)
        {
           all++;
           fre(j,1,mn+1)
             g[j].push_back(all);

          boy[i]=all;
          girl[all]=i;
        }
    }

   fre(i,mn+1,mn*2+1)
    if(girl[i]==0)
    {
        all++;

        fre(j,mn+1,mn*2+1)
         g[all].push_back(j);

        boy[all]=i;
        girl[i]=all;
    }

   fre(i,1,all+1)
   {
      if(boy[i])
        g[boy[i]].push_back(i);
   }
}

void tarjan(int x)
{
    int i,j;
    time[x]=low[x]=++time_num;

    instack[x]=1;
    q.push(x);

    fre(i,0,g[x].size())
    {
        int to=g[x][i];

        if(time[to]==0)
        {
            tarjan(to);

            if(low[to]<low[x])
                low[x]=low[to];
        }
       else
         if(instack[to]&&time[to]<low[x])
          low[x]=time[to];
    }

    int to;

    if(time[x]==low[x])
    {
        cnt++;

        do{
           to=q.top();
           q.pop();
            instack[to]=0;
           type[to]=cnt;
        }while(to!=x);

    }

}

void solve()
{
    int i,j;
    mem(time,0);
    mem(low,0);
    mem(instack,0);
    mem(type,0);

    cnt=time_num=0;

    while(!q.empty()) q.pop();

    fre(i,1,mn+1)
        if(time[i]==0)
            tarjan(i);

    int ans[N],k;

    fre(i,1,n+1)
    {
        k=0;

        fre(j,0,g[i].size())
        {
            int to=g[i][j];

            if(type[i]!=type[to]) continue;

            if(to-mn<=m)
                 ans[k++]=to-mn;


        }
        sort(ans,ans+k);

        pf("%d",k);
        fre(j,0,k)
          pf(" %d",ans[j]);
        pf("\n");
    }

}

int main()
{
    int i,j,t,ca=0;
    sf(t);
    while(t--)
    {
        sff(n,m);
        mn=max(n,m);

        fre(i,1,mn*4)
            g[i].clear();

        fre(i,1,n+1)
         {
            int k,x;
            sf(k);
            while(k--)
            {
                sf(x);
                g[i].push_back(x+mn);
            }
         }

       pf("Case #%d:\n",++ca);

       xiong();

       leave();
       solve();
    }
    return 0;
}

HDU 4685 Prince and Princess（二分图 + 强连通）

原文：http://blog.csdn.net/u014737310/article/details/44498997