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
用并查集合并女生的关系，再用二分跑最大流，因为次数具有单调性，所以可以二分
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];//注意这里要开的够大。。不然WA在这里真的想骂人。。问题是还不报RE。。
22 int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
23 int cnt=0;//实际存储总边数
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)//加边
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)//其实这个bfs可以融合到下面的迭代里，但是好像是时间要长
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);//从汇点开始反向建层次图
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)//注意是从汇点反向bfs，但应该判断正向弧的余量
59 {
60 dep[v]=dep[u]+1;
61 gap[dep[v]]++;
62 q.push(v);
63 //if(v==sp)//感觉这两句优化加了一般没错，但是有的题可能会错，所以还是注释出来，到时候视情况而定
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)//迭代寻找增广路,n为节点数
81 {
82 if(u==t)
83 {
84 int f=INF;
85 for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增广路
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;//当前弧优化
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优化
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;//常数优化
121 }
122 }
123 dep[u]=x+1;
124 gap[dep[u]]++;
125 if(u!=s)//当前点没有增广路则后退一个点
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 }
原文地址：https://www.cnblogs.com/Fighting-sh/p/9960178.html