HDU1281+二分图

一开始TLE了。。

/*
dfs+TLE
题意：给定一些点，求最多能放多少个点，使得这些放的点不互相攻击

*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>
#include<map>
#include<math.h>
typedef long long ll;
//typedef __int64 int64;
const int maxn = 105;
const int maxm = 1005;
const int inf = 0x7FFFFFFF;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int row[ maxn ],col[ maxn ];//该行该列能否放车
int ans_num;
struct node2{
    int x,y;
}Node[ 10005 ];//存储能放车的点
int CNT[ maxn ][ maxn ];

struct node{
    int x[ 24 ],y[ 24 ];
}solve[ maxm ];//每个方案数的具体方法

int cnt ;//记录有ansnum个车的方案数

void init(){
    //ans_num = 0;
    memset( row,0,sizeof( row ) );
    memset( col,0,sizeof( col ) );
}

void dfs( int now,int k,int sum ){
    for( int i=now;i<k;i++ ){
        if( row[ Node[i].x ]==0&&col[ Node[i].y ]==0 ){
            row[ Node[i].x ] = 1;
            col[ Node[i].y ] = 1;
            if( sum+1>ans_num ) ans_num = sum+1;
            //printf("sum1 = %d\n",sum);
            dfs( now+1,k,sum+1 );
            row[ Node[i].x ] = 0;
            col[ Node[i].y ] = 0;
        }
    }
    return ;
}

void dfs2( int now,int k,int sum ){
    if( sum==ans_num ){
        //printf("sum = %d\n",sum);
        int f = -1;
        for( int j=0;j<sum;j++ ){
            if( solve[cnt].x[j]==-1||solve[cnt].y[j]==-1 ) {
                f = 1;
                break;
            }
            //CNT[ solve[cnt].x[j] ][ solve[cnt].y[j] ] ++;
            //printf("%d %d\n",solve[cnt].x[j],solve[cnt].y[j]);
        }
        if( f==-1 ) {
            
            //printf("sum = %d\n",sum);
            for( int j=0;j<sum;j++ ){
                CNT[ solve[cnt].x[j] ][ solve[cnt].y[j] ] ++;
                //printf("%d %d\n",solve[cnt].x[j],solve[cnt].y[j]);
            }
            
            cnt++ ;
        }
        return ;
    }
    for( int i=now;i<k;i++ ){
        if( row[ Node[i].x ]==0&&col[ Node[i].y ]==0 ){
            row[ Node[i].x ] = 1;
            col[ Node[i].y ] = 1;
            int tmp = cnt;
            solve[ tmp ].x[ sum ] = Node[i].x;
            solve[ tmp ].y[ sum ] = Node[i].y;
            //printf("sum2 = %d\n",sum+1);
            dfs2( now+1,k,sum+1 );
            solve[ tmp ].x[ sum ] = -1;
            solve[ tmp ].y[ sum ] = -1;
            row[ Node[i].x ] = 0;
            col[ Node[i].y ] = 0;
        }
    }
    return ;
}

int main(){
    int n,m,k;
    int ca = 1;
    while( scanf("%d%d%d",&n,&m,&k)==3 ){
        int a,b;
        for( int i=0;i<k;i++ ){
            scanf("%d%d",&Node[i].x,&Node[i].y);
        }
        init();
        ans_num = 0;
        dfs( 0,k,0 );//求出最多能放多少个车
        if( ans_num==0 ){
            printf("Board %d have 0 important blanks for 0 chessmen.\n",ca++);
            continue;
        }
        //printf("ans_sum:%d\n\n\n",ans_num);
        cnt = 0;//统计放ansnum个车 有多少种方案
        init();
        memset( CNT,0,sizeof( CNT ) );
        for( int i=0;i<maxm;i++ ){
            for( int j=0;j<24;j++ )
                solve[i].x[j]=solve[i].y[j]=-1;
        }
        dfs2( 0,k,0 );
        int res = 0;
        for( int i=0;i<k;i++ ){
            if( CNT[Node[i].x][Node[i].y]==cnt ) res++;
            /*printf("CNT[%d][%d] = %d\n",Node[i].x,Node[i].y,CNT[Node[i].x][Node[i].y]);*/
        }
        printf("Board %d have %d important blanks for %d chessmen.\n",ca++,res,ans_num);
    }
    return 0;
}

二分图：

行号和列号进行匹配！！！！！！！！！！

/*
二分图
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>
#include<map>
#include<math.h>
typedef long long ll;
//typedef __int64 int64;
const int maxn = 105;
const int maxm = 1005;
const int inf = 0x7FFFFFFF;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int mat[ maxn ][ maxn ];
int mylink[ maxn ],vis[ maxn ];

int ans1 ,ans2;

bool dfs( int u,int n,int m ){
    for( int i=1;i<=m;i++ ){
        if( vis[ i ] ==-1&&mat[u][i]==1 ){
            vis[ i ] = 1;
            if( mylink[ i ]==-1||dfs( mylink[i],n,m ) ){
                mylink[ i ] = u;
                return true;
            }
        }
    }
    return false;
}

int max_match( int n,int m ){
    memset( mylink,-1,sizeof( mylink ) );
    int mylink2[ maxn ];
    int sum = 0;
    for( int i=1;i<=n;i++ ){
        memset( vis,-1,sizeof( vis ));
        if( dfs( i,n,m )==true )
            sum++;
    }
    for( int i=0;i<maxn;i++ ){
        mylink2[ i ] = mylink[ i ];
    }
    
    for( int i=1;i<=m;i++ ){
        if( mylink2[i]!=-1 ){
            int tx,ty;
            tx = mylink2[ i ];
            ty = i;
            mat[ tx ][ ty ] = 0;
            memset( mylink,-1,sizeof( mylink ) );
            int tmp_sum = 0;
            for( int i=1;i<=n;i++ ){
                memset( vis,-1,sizeof( vis ));
                if( dfs( i,n,m )==true )
                    tmp_sum++;
            }
            if( tmp_sum!=sum ) ans1++;
            mat[ tx ][ ty ] = 1;
        }
    }
    //对于已经选定的一种放车方案，分别 去掉 这种方案中的某个棋子，再看能否得到答案
    return sum;
}
 
int main(){
    int n,m,k;
    int ca = 1;
    while( scanf("%d%d%d",&n,&m,&k)==3 ){
        memset( mat,0,sizeof( mat ));
        while( k-- ){
            int a,b;
            scanf("%d%d",&a,&b);
            mat[ a ][ b ] = 1;
        }
        ans1 = 0;
        ans2 = max_match( n,m );
        printf("Board %d have %d important blanks for %d chessmen.\n",ca++,ans1,ans2);
    }
    return 0;
}