HDU 4687 Boke and Tsukkomi （一般图匹配带花树）

Boke and Tsukkomi

Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 336    Accepted Submission(s): 116

Problem Description

A new season of Touhou M-1 Grand Prix is approaching. Girls in Gensokyo cannot wait for participating it. Before the registration, they have to decide which combination they are going to compete as. Every girl in Gensokyo is both a boke (funny girl) and a tsukkomi (straight girl). Every candidate combination is made up of two girls, a boke and a tsukkomi. A girl may belong to zero or more candidate combinations, but one can only register as a member of one formal combination. The host of Touhou M-1 Grand Prix hopes that as many formal combinations as possible can participate in this year. Under these constraints, some candidate combinations are actually redundant as it\'s impossible to register it as a formal one as long as the number of formal combinations has to be maximized. So they want to figure out these redundant combinations and stop considering about them.

 

Input

There are multiple test cases. Process to the End of File.
The first line of each test case contains two integers: 1 ≤ N ≤ 40 and 1 ≤ M ≤ 123, where N is the number of girls in Gensokyo, and M is the number of candidate combinations. The following M lines are M candidate combinations, one by each line. Each combination is represented by two integers, the index of the boke girl 1 ≤ B_i ≤ N and the index of the tsukkomi girl 1 ≤ T_i ≤ N, where B_i != T_i.

 

Output

For each test case, output the number of redundant combinations in the first line. Then output the space-separated indexes of the redundant combinations in ascending order in the second line.

 

Sample Input

4 4
1 3
2 3
2 4
3 1
6 6
1 2
3 2
3 4
5 2
5 4
5 6

 

Sample Output

1
2
3
2 4 5

 

Author

Zejun Wu (watashi)

 

Source

2013 Multi-University Training Contest 9

 

Recommend

zhuyuanchen520

 

 

 

 

 

 

 

 

基本上就是一般图匹配的模板题了。

 

 

先一开始算总的匹配对数为cnt0.

然后枚举每一对匹配，把该对的点去掉，看匹配数是不是小于cnt0-1,是就是多余的。

 

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013/8/23 19:28:08
 File Name     :F:\2013ACM练习\专题学习\图论\一般图匹配带花树\HDU4687.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;
 const int MAXN = ;
 int N; //点的个数，点的编号从1到N
 bool Graph[MAXN][MAXN];
 int Match[MAXN];
 bool InQueue[MAXN],InPath[MAXN],InBlossom[MAXN];
 int Head,Tail;
 int Queue[MAXN];
 int Start,Finish;
 int NewBase;
 int Father[MAXN],Base[MAXN];
 int Count;
 void Push(int u)
 {
     Queue[Tail] = u;
     Tail++;
     InQueue[u] = true;
 }
 int Pop()
 {
     int res = Queue[Head];
     Head++;
     return res;
 }
 int FindCommonAncestor(int u,int v)
 {
     memset(InPath,false,sizeof(InPath));
     while(true)
     {
         u = Base[u];
         InPath[u] = true;
         if(u == Start) break;
         u = Father[Match[u]];
     }
     while(true)
     {
         v = Base[v];
         if(InPath[v])break;
         v = Father[Match[v]];
     }
     return v;
 }
 void ResetTrace(int u)
 {
     int v;
     while(Base[u] != NewBase)
     {
         v = Match[u];
         InBlossom[Base[u]] = InBlossom[Base[v]] = true;
         u = Father[v];
         if(Base[u] != NewBase) Father[u] = v;
     }
 }
 void BloosomContract(int u,int v)
 {
     NewBase = FindCommonAncestor(u,v);
     memset(InBlossom,false,sizeof(InBlossom));
     ResetTrace(u);
     ResetTrace(v);
     if(Base[u] != NewBase) Father[u] = v;
     if(Base[v] != NewBase) Father[v] = u;
     for(int tu = ; tu <= N; tu++)
         if(InBlossom[Base[tu]])
         {
             Base[tu] = NewBase;
             if(!InQueue[tu]) Push(tu);
         }
 }
 void FindAugmentingPath()
 {
     memset(InQueue,false,sizeof(InQueue));
     memset(Father,,sizeof(Father));
     for(int i = ;i <= N;i++)
         Base[i] = i;
     Head = Tail = ;
     Push(Start);
     Finish = ;
     while(Head < Tail)
     {
         int u = Pop();
         for(int v = ; v <= N; v++)
             if(Graph[u][v] && (Base[u] != Base[v]) && (Match[u] != v))
             {
                 if((v == Start) || ((Match[v] > ) && Father[Match[v]] > ))
                     BloosomContract(u,v);
                 else if(Father[v] == )
                 {
                     Father[v] = u;
                     if(Match[v] > )
                         Push(Match[v]);
                     else
                     {
                         Finish = v;
                         return;
                     }
                 }
             }
     }
 }
 void AugmentPath()
 {
     int u,v,w;
     u = Finish;
     while(u > )
     {
         v = Father[u];
         w = Match[v];
         Match[v] = u;
         Match[u] = v;
         u = w;
     }
 }
 void Edmonds()
 {
     memset(Match,,sizeof(Match));
     for(int u = ; u <= N; u++)
         if(Match[u] == )
         {
             Start = u;
             FindAugmentingPath();
             if(Finish > )AugmentPath();
         }
 }
 int getMatch()
 {
     Edmonds();
     Count = ;
     for(int u = ; u <= N;u++)
         if(Match[u] > )
             Count++;
     return Count/;
 }

 bool g[MAXN][MAXN];
 pair<int,int>p[];
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int m;
     while(scanf("%d%d",&N,&m)==)
     {
         memset(g,false,sizeof(g));
         memset(Graph,false,sizeof(Graph));
         int u,v;
         for(int i = ;i <= m;i++)
         {
             scanf("%d%d",&u,&v);
             p[i] = make_pair(u,v);
             g[u][v] = true;
             g[v][u] = true;
             Graph[u][v] = true;
             Graph[v][u] = true;
         }
         int cnt0 = getMatch();
         //cout<<cnt0<<endl;
         vector<int>ans;
         for(int i = ;i <= m;i++)
         {
             u = p[i].first;
             v = p[i].second;
             memcpy(Graph,g,sizeof(g));
             for(int j = ;j <= N;j++)
                 Graph[j][u] = Graph[u][j] = Graph[j][v] = Graph[v][j] = false;
             int cnt = getMatch();
             //cout<<cnt<<endl;
             if(cnt < cnt0-)
                 ans.push_back(i);
         }
         int sz = ans.size();
         printf("%d\n",sz);
         for(int i = ;i < sz;i++)
         {
             printf("%d",ans[i]);
             if(i < sz-)printf(" ");
         }
         printf("\n");
     }
     return ;
 }