POJ 2446 Chessboard

Chessboardtime limit:2000msmemory limit:65536kbthis problem'll be judged onPKU. Original id:2446
64-bit integer IO format: %lld Java class name: Main 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 have to follow the rules below:
1. Any normal grid should is covered with exactly one card.
2. One card should cover exactly 2 normal adjacent grids.

Some examples is given in the figures below:

A VALID solution.

An invalid solution, because the hole of red color are covered with a card.

An invalid solution, because there exists a grid, which are not covered.
Your task is to help Bob to decide whether or not the chessboard can being covered according to the rules above.InputThere is 3 integers in the first line:m, N, K (0 < M, n <=, 0 <= k < m * N), the number of rows, column A nd holes. In the next k lines, there are a pair of integers (x, y) in each line, which represents a hole in the y-th row, the X-th Co Lumn.OutputIf 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.SourcePOJ monthly,charlescpp Problem solving: Maximum match

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5 using namespacestd;6 Const intMAXN = -* -;7 BOOLmp[maxn][maxn],used[maxn],cant[ -][ -];8 intlink[maxn],n,m,k;9 BOOLMatchintu) {Ten      for(inti =0; i < MAXN; ++i) { One         if(!mp[u][i] | | used[i])Continue; AUsed[i] =true; -         if(Link[i] = =-1||match (Link[i])) { -Link[i] =u; the             return true; -         } -     } -     return false; + } - intMain () { +     intx, y; A      while(~SCANF ("%d%d%d",&n,&m,&k)) { atmemset (Cant,false,sizeofcant); -          for(inti =0; I < K; ++i) { -scanf"%d%d",&y,&x); -cant[x-1][y-1] =true; -         } -Memset (MP,false,sizeofMP); in          for(inti =0; I < n; ++i) { -              for(intj =0; J < M; ++j) { to                 if((I+J) &1) &&!Cant[i][j]) { +                     if(I >0&&!cant[i-1][J]) Mp[i*m + j][(i-1) *m + j] =true; -                     if(J >0&&!cant[i][j-1]) Mp[i*m + j][i*m + J-1] =true; the                     if(i +1< n &&!cant[i+1][J]) Mp[i*m + j][(i+1) *m + j] =true; *                     if(j +1< M &&!cant[i][j+1]) Mp[i*m + j][i*m + j +1] =true; $                 }Panax Notoginseng             } -         } the         intRET =0; +memset (link,-1,sizeofLink); A          for(inti =0; I <= m*n; ++i) { thememset (Used,false,sizeofused); +             if(Match (i)) + +ret; -         } $printf"%s\n", n*m-k = =2*ret?"YES":"NO"); $     } -     return 0; -}

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POJ 2446 Chessboard