Marriage Match II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2507 Accepted Submission(s): 856
Problem Description
Presumably, you all have known the question of stable marriage match. A girl will choose a boy; it is similar as the game of playing house we used to play when we are kids. What a happy time as so many friends playing 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. 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.
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 a integer T, means the number of test cases.
Each test case starts with three integer n, m and f in a line (3<=n<=100,0<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 2 1 1 2 3 3 2 4 2 4 4 1 4 2 3
Sample Output
2
Author
starvae
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。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。
解析：对于女的与女的之间关系可以用并查集处理一下就OK 了，关键是如何得到多少组配对情况，那么分析一下，根据题意，每个人在每一组配对情况就是不相同的，那么如果有k组，则所有女性每人至少有k个不同的配对关系。同样，对于所有的男性配对女性也是同样的，至少有k个配对关系。就可以用最大流来做，女与源点相连，男与汇点相连的边容都为k，女与男配对的关系因为只能用一次，所以边容设为1。如果得到的最大流为 n*k 的值则成立。对于怎么枚举出k，则用二分答案。
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
#define captype int
const int MAXN = 210; //点的总数
const int MAXM = 40010; //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]; //每种距离(或可认为是高度)点的个数
int dis[MAXN]; //每个点到终点eNode 的最短距离
int cur[MAXN]; //cur[u] 表示从u点出发可流经 cur[u] 号边
int pre[MAXN];
void init(){
eid=0;
memset(head,-1,sizeof(head));
}
//有向边 三个参数，无向边4个参数
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 sNode,int eNode , int n){
memset(gap,0,sizeof(gap));
memset(dis,0,sizeof(dis));
memcpy(cur,head,sizeof(head));
pre[sNode]=-1;
gap[0]=n;
captype ans = 0;
int u=sNode;
while(dis[sNode]<n){
if(u==eNode){
captype mint = INF , mincap = INF;
int minid ;
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 ;
minid=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[minid^1].to;
continue;
}
bool flag=false;
for(int i=cur[u]; i!=-1; i=edg[i].next)
if(edg[i].cap-edg[i].flow>0 && 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>0 && minh>dis[edg[i].to]){
minh=dis[edg[i].to];
cur[u]=i;
}
gap[dis[u]]--;
if(gap[dis[u]]==0)
return ans;
dis[u]=minh+1;
gap[dis[u]]++;
if(u!=sNode)
u=edg[pre[u]^1].to;
}
return ans;
}
void changCap(int s,int t,int k){
for(int i=head[s]; i!=-1; i=edg[i].next)
edg[i].cap=k;
for(int i=head[t]; i!=-1; i=edg[i].next)
edg[i^1].cap=k;
for(int i=0; i<eid; i++)//每一次都要清0
edg[i].flow=0;
}
int father[MAXN];
int findfath(int x){
if(x!=father[x])
father[x]=findfath(father[x]);
return father[x];
}
void link(int x,int y){
x=findfath(x);
y=findfath(y);
father[x]=y;
}
int mapt[105][105];
void changMap(int n){
int tmp[105][105]={0};
for(int i=1; i<=n; i++) //对于在同一棵树上可以用一个点来替换（根节点）
father[i]=findfath(i);
for(int i=1; i<=n; i++){
int ti=father[i];
for(int j=1; j<=n; j++)
tmp[ti][j]|=mapt[i][j];
}
for(int i=1; i<=n; i++){
int ti=father[i];
for(int j=1; j<=n; j++)
mapt[i][j]=tmp[ti][j];
}
}
int main(){
int T;
int n,m,f;
int a,b;
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&n,&m,&f);
memset(mapt,0,sizeof(mapt));
while(m--){
scanf("%d%d",&a,&b);
mapt[a][b]=1;
}
for(int i=1; i<=n; i++)
father[i]=i;
while(f--){
scanf("%d%d",&a,&b);
link(a,b);
}
changMap(n);
init();
int s=0 , t=2*n+1;
//女用1~n点表示，男用n+1~n+n表示
for(int i=1; i<=n; i++){
addEdg(s,i,0);
addEdg(i+n,t,0);
for(int j=1; j<=n; j++)
if(mapt[i][j])
addEdg(i,j+n,1);
}
int l=0,r=n ,ans=0;
while(l<=r){
m=(l+r)>>1;
changCap(s,t,m);
int maxflow=maxFlow_sap(s,t,t+1);
if(maxflow==n*m)
ans=m , l=m+1;
else r=m-1;
}
printf("%d\n",ans);
}
}