HDU 3277 Marriage Match III(并查集+二分答案+最大流SAP)拆点，经典

Marriage Match III

Time Limit: 10000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1581 Accepted Submission(s): 464



Problem Description
Presumably, you all have known the question of stable marriage match. A girl will choose a boy; it is similar as the ever game of play-house . What a happy time as so many friends play together. And it is normal that a fight or a
quarrel breaks out, but we will still play together after that, because we are kids.

Now, there are 2n kids, n boys numbered from 1 to n, and n girls numbered from 1 to n. As you know, ladies first. So, every girl can choose a boy first, with whom she has not quarreled, to make up a family. Besides, the girl X can also choose boy Z to be her
boyfriend when her friend, girl Y has not quarreled with him. Furthermore, the friendship is mutual, which means a and c are friends provided that a and b are friends and b and c are friend.

Once every girl finds their boyfriends they will start a new round of this game—marriage match. At the end of each round, every girl will start to find a new boyfriend, who she has not chosen before. So the game goes on and on. On the other hand, in order to
play more times of marriage match, every girl can accept any K boys. If a girl chooses a boy, the boy must accept her unconditionally whether they had quarreled before or not.

Now, here is the question for you, how many rounds can these 2n kids totally play this game?



Input
There are several test cases. First is an integer T, means the number of test cases.

Each test case starts with three integer n, m, K and f in a line (3<=n<=250, 0<m<n*n, 0<=f<n). n means there are 2*n children, n girls(number from 1 to n) and n boys(number from 1 to n).

Then m lines follow. Each line contains two numbers a and b, means girl a and boy b had never quarreled with each other.

Then f lines follow. Each line contains two numbers c and d, means girl c and girl d are good friends.



Output
For each case, output a number in one line. The maximal number of Marriage Match the children can play.


Sample Input

1
4 5 1 2
1 1
2 3
3 2
4 2
4 4
1 4
2 3

Sample Output

3

Author
starvae


Source
HDOJ Monthly Contest – 2010.01.02

题意：共有2*n个人，一半女一半男，女与男有m个关系，表示可以成为一对，接下来 f 对女的与女的 的朋友关系，如果a与b是朋友，那么表示a女与b女的相连男性也可以成为一对，同样b也与a的相连男性可成为一对，女的之间的朋友关系可以传递，且每个女性可以再任意选择K人。一组配对情况为所有的女性都有一个与之配对的男性（一对一的关系），如果还有其他组配对情况，那么所有的女性配对不可以再与原来的男性配成对。问最多有多少组配对情况。

这题和HDU3081 很类似。

但是因为可以随意选择K个人。

所以要将女孩拆成两个点。

将每个女孩u分为u1，u2，若u喜欢v则加一条u1到v的边 否则加一条u2到v的边，令加u1到u2的容量为k的边；

这个拆点的想法非常巧妙。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define captype int

const int MAXN = 100010;   //点的总数
const int MAXM = 4000100;    //边的总数
const int INF = 1<<30;
struct EDG{
    int to,next;
    captype cap,flow;
}edg[MAXM];
int eid,head[MAXN];
int gap[MAXN];
int dis[MAXN];
int cur[MAXN];
int pre[MAXN];

void init(){
    eid=0;
    memset(head,-1,sizeof(head));
}
void addEdg(int u,int v,captype c,captype rc=0){
    edg[eid].to=v; edg[eid].next=head[u];
    edg[eid].cap=c; edg[eid].flow=0; head[u]=eid++;

    edg[eid].to=u; edg[eid].next=head[v];
    edg[eid].cap=rc; edg[eid].flow=0; head[v]=eid++;
}
captype maxFlow_sap(int s,int t,int n){
    memset(gap,0,sizeof(gap));
    memset(dis,0,sizeof(dis));
    memcpy(cur,head,sizeof(head));
    pre[s]=-1;
    gap[0]=n;

    captype ans=0;
    int u=s;
    while(dis[s]<n){
        if(u==t){
            captype mint=INF;
            int id;
            for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to])
            if(mint>edg[i].cap-edg[i].flow){
                mint=edg[i].cap-edg[i].flow;
                id=i;
            }
            for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to]){
                edg[i].flow+=mint;
                edg[i^1].flow-=mint;
            }
            ans+=mint;
            u=edg[id^1].to;
            continue;
        }
        bool flag=0;
        for(int i=cur[u]; i!=-1; i=edg[i].next)
        if(edg[i].cap>edg[i].flow&&dis[u]==dis[edg[i].to]+1){
            cur[u]=pre[edg[i].to]=i;
            flag=true;
            break;
        }
        if(flag){
            u=edg[cur[u]].to;
            continue;
        }
        int minh=n;
        for(int i=head[u]; i!=-1; i=edg[i].next)
        if(edg[i].cap>edg[i].flow && minh>dis[edg[i].to]){
            cur[u]=i; minh=dis[edg[i].to];
        }
        gap[dis[u]]--;
        if(!gap[dis[u]]) return ans;
        dis[u]=minh+1;
        gap[dis[u]]++;
        if(u!=s)
         u=edg[pre[u]^1].to;
    }
    return ans;
}

int fath[MAXN];
int findroot(int x){
    if(x!=fath[x])
     fath[x]=findroot(fath[x]);
     return fath[x];
}
void setroot(int x,int y){
    x=findroot(x);
    y=findroot(y);
    fath[x]=y;
}
void rebuildMap(int mapt[255][255],int n){//处理朋友之间的关系
    int mp[255][255]={0};
    for(int i=1; i<=n; i++)
    fath[i]=findroot(i);
    for(int i=1; i<=n; i++){
        int j=fath[i];
        for(int e=1; e<=n; e++)
        mp[j][e]|=mapt[i][e];
    }
    for(int i=1; i<=n; i++){
        int j=fath[i];
        for(int e=1; e<=n; e++)
        mapt[i][e]=mp[j][e];
    }
}
int main()
{
    int T,n,m,k,f,mapt[255][255];
    int u,v;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d%d",&n,&m,&k,&f);

        init();
        memset(mapt,0,sizeof(mapt));
        for(int i=1; i<=n; i++)
            fath[i]=i;

        while(m--){
            scanf("%d%d",&u,&v);
            mapt[u][v]=1;
        }
        while(f--){
            scanf("%d%d",&u,&v);
            setroot(u,v);
        }
        rebuildMap(mapt,n);

        int s=0, t=3*n+1;
        for(int i=1; i<=n; i++){
            addEdg(s,i,0);
            addEdg(i,i+n,k);
            for(int j=1; j<=n; j++)
            if(mapt[i][j])
                addEdg(i,j+2*n,1);
            else
                addEdg(i+n,j+2*n,1);

            addEdg(i+2*n,t,0);
        }
        
        int ans=0 , l=0 , r=n ,mid;
        while(l<=r){
            mid=(l+r)>>1;
            
            for(int i=0; i<eid; i++)
                edg[i].flow=0;
            for(int i=head[s]; i!=-1; i=edg[i].next)
                edg[i].cap=mid;
            for(int i=head[t]; i!=-1; i=edg[i].next)
                edg[i^1].cap=mid;
                
            if(n*mid==maxFlow_sap(s,t,t+1))
                ans=mid,l=mid+1;
            else
                r=mid-1;
        }
        
        printf("%d\n",ans);
    }
}