poj2446--Chessboard(二分匹配)

Chessboard

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 12811 Accepted: 4005

Description

Alice and Bob often play games on chessboard. One day, Alice draws a board with size M * N. She wants Bob to use a lot of cards with size 1 * 2 to cover the board. However, she thinks it too easy to bob, so she makes some holes on the board (as shown in the figure below). 

We call a grid, which doesn’t contain a hole, a normal grid. Bob has to follow the rules below: 
1. Any normal grid should be covered with exactly one card. 
2. One card should cover exactly 2 normal adjacent grids. 

Some examples are given in the figures below: 
 
A VALID solution.
 
An invalid solution, because the hole of red color is covered with a card.
 
An invalid solution, because there exists a grid, which is not covered.
Your task is to help Bob to decide whether or not the chessboard can be covered according to the rules above.

Input

There are 3 integers in the first line: m, n, k (0 < m, n <= 32, 0 <= K < m * n), the number of rows, column and holes. In the next k lines, there is a pair of integers (x, y) in each line, which represents a hole in the y-th row, the x-th column.

Output

If the board can be covered, output "YES". Otherwise, output "NO".

Sample Input

4 3 22 13 3

Sample Output

YES

Hint

 
A possible solution for the sample input.

Source

POJ Monthly,charlescpp

使用1*2的小长方块覆盖n*m的区域，其中给出的t个点不能覆盖，问存不存该可能

二分匹配的题目，建图是最关键的，1*2的小方块会覆盖住两个方格，所以相邻的方块一定是在不同的点集中，我用的m[][]数组记录n*m，用bfs来分配点，得到左右点集，如果左右点集中的点相邻，就意味着他们之间有一条线，对这个图做二分匹配，求出最大匹配，如果存在由1*2的方块覆盖整体面积，那么就会有  n*m-t = 最大匹配数 * 2 ，否则不存在全部覆盖的可能

#include <stdio.h>#include <string.h>#include <math.h>int p[40][40] ;//0是空白点，-1不可选点，1左点集，2右点集struct node1{    int x , y ;} l[1600];struct node2{    int x , y ;} r[1600];struct node{    int x , y ;} d[1600] ;int nl , nr , n , m ;int link[1600] , temp[1600] ;int f(int x){    int i , j ;    for(i = 0 ; i < nr ; i++)    {        if( temp[i] == 0 &&  fabs(l[x].x - r[i].x) + fabs( l[x].y - r[i].y ) == 1  )        {            temp[i] = 1 ;            if( link[i] == -1 || f( link[i] ) )            {                link[i] = x ;                return 1 ;            }        }    }    return 0 ;}int dd[][2] = { {0,-1},{0,1},{1,0},{-1,0} } ;void bfs(int i,int j){    int low = 0 , top = 1 , u , v ;    d[0].x = i ;    d[0].y = j ;    p[i][j] = 1 ;    while(low < top)    {        node mid = d[low++] ;        for(i = 0 ; i < 4 ; i++)        {            u = mid.x + dd[i][0] ;            v = mid.y + dd[i][1] ;            if(u > 0 && u <= n && v > 0 && v <= m && p[u][v] == 0 )            {                if( p[mid.x][mid.y] == 1 )                    p[u][v] = 2 ;                else                    p[u][v] = 1 ;                d[top].x = u ;                d[top++].y = v ;            }        }    }}int main(){    int t , tt , i , j , ans ;    while(scanf("%d %d %d", &n, &m, &t)!=EOF)    {        tt = t ;        memset(p,0,sizeof(p));        memset(link,-1,sizeof(link));        nl = 0 ;        nr = 0 ;        ans = 0 ;        while(t--)        {            scanf("%d %d", &j, &i);            p[i][j] = -1 ;        }        if( (n*m-tt)%2 )        {            printf("NO\n");            continue ;        }        for(i = 1 ; i <= n ; i++)            for(j = 1 ; j <= m ; j++)            {                if( p[i][j] == 0 )                {                    bfs(i,j);                }            }        for(i = 1 ; i <= n ; i++)            for(j = 1 ; j <= m ; j++)            {                if(p[i][j] == 1)                {                    l[nl].x = i ;                    l[nl++].y = j ;                }                else if(p[i][j] == 2)                {                    r[nr].x = i ;                    r[nr++].y = j ;                }            }        for(i = 0 ; i < nl ; i++)        {            memset(temp,0,sizeof(temp));            if( f(i) )            {                ans++ ;            }        }        if( (n*m-tt) == ans*2 )            printf("YES\n");        else            printf("NO\n");    }    return 0;}