简单二分图匹配

1.HDU 2063 过山车

模板题。。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <map>
#include <set>
#include <stdlib.h>
#include <time.h>

using namespace std;


int color[1100];
int match[1100];
int mp[1100][1100];
int leftnum, rightnum;
 

void init( )
{
     for( int i = 1; i <= 1000; i++) {
          color[i] =  0;
          match[i] = -1;
     }
     for( int i = 1; i <= 1000; i++) {
          for( int j = 1; j <= 1000; j++)
               mp[i][j] = 0;
     }
} 

int find( int x)
{
  for( int i = 1; i <= rightnum; i++) {
         if (!color[i] && mp[x][i]) {     
            color[i] = 1;
            if (match[i] == -1 || find(match[i])) {
                match[i] = x;
                return 1;
            }
        }
  }
    return 0;
}


int solve( )
{ 
  int cnt = 0;
  for( int  i = 1; i <= leftnum; i++) {
       memset(color, 0, sizeof(color));
       if ( find(i) ) cnt++;
  }
  return cnt;
}

int main( )
{ 
  int k, M, N, a, b;
  
  while ( scanf("%d", &k), k) {
        scanf("%d%d", &M, &N);
        init( );
        leftnum = rightnum = 0;
        for( int i = 1; i <= k; i++)
        {
          scanf("%d%d", &a, &b);
          mp[a][b] = 1;
          if (a > leftnum )
             leftnum = a;
          if (b > rightnum )
             rightnum = b;
        }
        printf("%d\n", solve( )); 
  }
  return 0;    
}

2.HDU 2444

要判断能不能构成二分图。。构好图。

然后就直接用匈牙利算法就可以了

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int left[1000], right[1000], ld, rd;
int mp[300][300];
int match[300];
int color[300];

void init( )
{
  memset(left, 0, sizeof(left));
  memset(right, 0, sizeof(right));
  memset(match, -1, sizeof(match));
  memset(mp, 0, sizeof(mp));
  ld = 0, rd = 0;
    
}

int find( int x )
{
  for( int j = 1; j < rd; j++)
  {
    int i = right[j];
    //printf("color = %d mp = %d\n",mp[x][i], color[i]);
    if( mp[x][i] && !color[i] )
    {
      color[i] = 1;
      //puts("A");
      if( match[i] == -1 || find(match[i]))
      {
         match[i] = x;
         return 1;
      }
    }
  
  }
  return 0;
}

void solve(  )
{
  int cnt = 0;
  for( int j = 1; j < ld; j++)
  {
     int i = left[j];
     memset(color, 0, sizeof(color));
     if( find(i) )
        cnt++;
  }
  printf("%d\n", cnt);
}
int main( )
{
   int N, M, a, b;
   while( scanf("%d%d",&N, &M) != EOF )
   {
     init( );
     int x = 0;
     for( int i = 1; i <= M; i++)
     {
        scanf("%d%d",&a,&b);
        mp[a][b] = mp[b][a] = 1;
        int f1 = 0, f2 = 0;
        if( !left[a] && !right[a])
            left[a] = 1, f1 = 1, ld++;
        if( !left[a] && !right[b] && !f1)
            left[b] = 1, f2 = 1, ld++;
        if( !left[a] && !right[a] && !f1)
            right[a] = 1, rd++;
        if( !left[b] && !right[b] && !f2)
            right[b] = 1, rd++;
         if( (left[a] && left[b]) || (right[a] && right[b] ))
              x = 1;
     }
    // printf("%d \n", x);
     if( x == 1 || ld == N || rd == N )
     {
        puts("No");
        continue;
     }
     ld = rd = 1;
     //X集合  
     for( int i = 1; i <= N; i++)
        if( left[i] )
        left[ld++] = i;
     //Y集合
     for( int i = 1; i <= N; i++)
        if( right[i] )
        right[rd++] = i; 
     solve( );
   }
}