Marriage Match II(二分+並查集+最大流，好題)

Marriage Match II

http://acm.hdu.edu.cn/showproblem.php?pid=3081

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5420    Accepted Submission(s): 1739

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

 

用並查集合並女生的關係，再用二分跑最大流，由於次數具備單調性，因此能夠二分

  1 #include<iostream>
  2 #include<cstring>
  3 #include<string>
  4 #include<cmath>
  5 #include<cstdio>
  6 #include<algorithm>
  7 #include<queue>
  8 #include<vector>
  9 #include<set>
 10 #define maxn 200005
 11 #define MAXN 200005
 12 #define mem(a,b) memset(a,b,sizeof(a))
 13 const int N=200005;
 14 const int M=200005;
 15 const int INF=0x3f3f3f3f;
 16 using namespace std;
 17 int n;
 18 struct Edge{
 19     int v,next;
 20     int cap,flow;
 21 }edge[MAXN*20];//注意這裏要開的夠大。。否則WA在這裏真的想罵人。。問題是還不報RE。。
 22 int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
 23 int cnt=0;//實際存儲總邊數
 24 void isap_init()
 25 {
 26     cnt=0;
 27     memset(pre,-1,sizeof(pre));
 28 }
 29 void isap_add(int u,int v,int w)//加邊
 30 {
 31     edge[cnt].v=v;
 32     edge[cnt].cap=w;
 33     edge[cnt].flow=0;
 34     edge[cnt].next=pre[u];
 35     pre[u]=cnt++;
 36 }
 37 void add(int u,int v,int w){
 38     isap_add(u,v,w);
 39     isap_add(v,u,0);
 40 }
 41 bool bfs(int s,int t)//其實這個bfs能夠融合到下面的迭代裏，可是好像是時間要長
 42 {
 43     memset(dep,-1,sizeof(dep));
 44     memset(gap,0,sizeof(gap));
 45     gap[0]=1;
 46     dep[t]=0;
 47     queue<int>q;
 48     while(!q.empty())
 49     q.pop();
 50     q.push(t);//從匯點開始反向建層次圖
 51     while(!q.empty())
 52     {
 53         int u=q.front();
 54         q.pop();
 55         for(int i=pre[u];i!=-1;i=edge[i].next)
 56         {
 57             int v=edge[i].v;
 58             if(dep[v]==-1&&edge[i^1].cap>edge[i^1].flow)//注意是從匯點反向bfs，但應該判斷正向弧的餘量
 59             {
 60                 dep[v]=dep[u]+1;
 61                 gap[dep[v]]++;
 62                 q.push(v);
 63                 //if(v==sp)//感受這兩句優化加了通常沒錯，可是有的題可能會錯，因此仍是註釋出來，到時候視狀況而定
 64                 //break;
 65             }
 66         }
 67     }
 68     return dep[s]!=-1;
 69 }
 70 int isap(int s,int t)
 71 {
 72     if(!bfs(s,t))
 73     return 0;
 74     memcpy(cur,pre,sizeof(pre));
 75     //for(int i=1;i<=n;i++)
 76     //cout<<"cur "<<cur[i]<<endl;
 77     int u=s;
 78     path[u]=-1;
 79     int ans=0;
 80     while(dep[s]<n)//迭代尋找增廣路,n爲節點數
 81     {
 82         if(u==t)
 83         {
 84             int f=INF;
 85             for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增廣路
 86                 f=min(f,edge[i].cap-edge[i].flow);
 87             for(int i=path[u];i!=-1;i=path[edge[i^1].v])
 88             {
 89                 edge[i].flow+=f;
 90                 edge[i^1].flow-=f;
 91             }
 92             ans+=f;
 93             u=s;
 94             continue;
 95         }
 96         bool flag=false;
 97         int v;
 98         for(int i=cur[u];i!=-1;i=edge[i].next)
 99         {
100             v=edge[i].v;
101             if(dep[v]+1==dep[u]&&edge[i].cap-edge[i].flow)
102             {
103                 cur[u]=path[v]=i;//當前弧優化
104                 flag=true;
105                 break;
106             }
107         }
108         if(flag)
109         {
110             u=v;
111             continue;
112         }
113         int x=n;
114         if(!(--gap[dep[u]]))return ans;//gap優化
115         for(int i=pre[u];i!=-1;i=edge[i].next)
116         {
117             if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x)
118             {
119                 x=dep[edge[i].v];
120                 cur[u]=i;//常數優化
121             }
122         }
123         dep[u]=x+1;
124         gap[dep[u]]++;
125         if(u!=s)//當前點沒有增廣路則後退一個點
126         u=edge[path[u]^1].v;
127      }
128      return ans;
129 }
130 
131 int m,d;
132 struct sair{
133     int x,y;
134 }p[maxn];
135 int Friend[205][205];
136 int fa[maxn];
137 
138 int Find(int x){
139     int r=x,y;
140     while(x!=fa[x]){
141         x=fa[x];
142     }
143     while(r!=x){
144         y=fa[r];
145         fa[r]=x;
146         r=y;
147     }
148     return x;
149 }
150 
151 void join(int x,int y){
152     int xx=Find(x);
153     int yy=Find(y);
154     if(xx!=yy){
155         fa[xx]=yy;
156     }
157 }
158 int tmp;
159 int Check(int mid){
160     isap_init();
161     int s=0,t=n+n+1;
162     for(int i=1;i<=n;i++){
163         for(int j=1;j<=n;j++){
164             if(Friend[i][j]){
165                 add(i,j+n,1);
166             }
167         }
168     }
169     for(int i=1;i<=n;i++){
170         add(s,i,mid);
171         add(n+i,t,mid);
172     }
173     n=n+n+2;
174     int tttt=isap(s,t);
175     n=tmp;
176     return tttt;
177 }
178 
179 int main(){
180     std::ios::sync_with_stdio(false);
181     int T;
182     cin>>T;
183     for(int co=1;co<=T;co++){
184         cin>>n>>m>>d;
185         tmp=n;
186         memset(Friend,0,sizeof(Friend));
187         for(int i=0;i<=n;i++) fa[i]=i;
188         for(int i=1;i<=m;i++) cin>>p[i].x>>p[i].y;
189         for(int i=m+1;i<=m+d;i++) cin>>p[i].x>>p[i].y;
190         for(int i=m+1;i<=m+d;i++) join(p[i].x,p[i].y);
191         for(int i=1;i<=m;i++){
192             for(int j=1;j<=n;j++){
193                 if(Find(p[i].x)==Find(j)&&!Friend[j][p[i].y]){
194                     Friend[j][p[i].y]=1;
195                 }
196             }
197         }
198         int L=0,R=n,mid;
199         while(L<=R){
200             mid=(L+R)>>1;
201             n=tmp;
202             if(Check(mid)>=(n*mid)){
203                 L=mid+1;
204             }
205             else{
206                 R=mid-1;
207             }
208         }
209         cout<<R<<endl;
210     }
211 }

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