hdu2458(二分图+最大独立数+匈牙利算法)

Kindergarten

[b]Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 490    Accepted Submission(s): 282
[/b]

[align=left]Problem Description[/align]
In a kindergarten, there are a lot of kids. All girls of the kids know each other and all boys also know each other. In addition to that, some girls and boys know each other. Now the teachers want to pick some kids to play a game,
which need that all players know each other. You are to help to find maximum number of kids the teacher can pick.
 

[align=left]Input[/align]
The input consists of multiple test cases. Each test case starts with a line containing three integers

G, B (1 ≤ G, B ≤ 200) and M (0 ≤ M ≤ G × B), which is the number of girls, the number of boys and

the number of pairs of girl and boy who know each other, respectively.

Each of the following M lines contains two integers X and Y (1 ≤ X≤ G,1 ≤ Y ≤ B), which indicates that girl X and boy Y know each other.

The girls are numbered from 1 to G and the boys are numbered from 1 to B.

The last test case is followed by a line containing three zeros.
 

[align=left]Output[/align]
For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the maximum number of kids the teacher can pick.
 

[align=left]Sample Input[/align]

2 3 3
1 1
1 2
2 3
2 3 5
1 1
1 2
2 1
2 2
2 3
0 0 0

 

[align=left]Sample Output[/align]

Case 1: 3
Case 2: 4 本题要求相互认识的人的最大个数，即求最大独立团的个数（即补图的最大独立数），有最大独立数=顶点数-最大点覆盖数，最大点覆盖数=最大匹配，最大匹配可以运用匈牙利算法求解 

#include<iostream>
#include<cstdio>
using namespace std;

const int MAXN=200+10;
int G,B;//u,v数目
int g[MAXN][MAXN];
int linker[MAXN];
bool visited[MAXN];

bool dfs(int u)//从左边开始找增广路径
{
int v;
for(v=1;v<=B;v++)//这个顶点编号从0开始，若要从1开始需要修改
if(g[u][v]&&!visited[v])
{
visited[v]=true;
if(linker[v]==-1||dfs(linker[v]))
{//找增广路，反向
linker[v]=u;
return true;
}
}
return false;//这个不要忘了，经常忘记这句
}

int hungary()
{
int res=0;
int u;
memset(linker,-1,sizeof(linker));
for(u=1;u<=G;u++)
{
memset(visited,0,sizeof(visited));
if(dfs(u)) res++;
}
return res;
}

int main()
{
int M,i,a,b,j;
int tag=1;
while(scanf("%d%d%d",&G,&B,&M))
{
if(0==G&&0==B&&0==M)
break;
memset(g,1,sizeof(g));

for(i=0;i<M;i++)
{
scanf("%d%d",&a,&b);
g[a][b]=0;
}

printf("Case %d: %d\n",tag++,G+B-hungary());
}
return 0;
}