poj 2446 Chessboard (二分图利用奇偶性匹配)
2014-07-31 21:29
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Chessboard
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
Sample Output
Hint
A possible solution for the sample input.
给你一个棋盘,棋盘上有几个洞,要求你用一些1*2的卡片覆盖没有洞的区域,一个格子只能有一张卡片覆盖。看是否能恰好覆盖。
二分匹配:利用二分匹配,两个能匹配的格子的坐标和必然奇偶性不同,利用这一点可以减少时间耗费。
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 13176 | Accepted: 4118 |
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 2 2 1 3 3
Sample Output
YES
Hint
A possible solution for the sample input.
给你一个棋盘,棋盘上有几个洞,要求你用一些1*2的卡片覆盖没有洞的区域,一个格子只能有一张卡片覆盖。看是否能恰好覆盖。
二分匹配:利用二分匹配,两个能匹配的格子的坐标和必然奇偶性不同,利用这一点可以减少时间耗费。
#include"stdio.h" #include"string.h" #include"queue" using namespace std; #define N 35 #define M 1200 int g ,n,m; int dir[4][2]={0,1,0,-1,-1,0,1,0}; int mark[M],link[M]; int judge(int x,int y) { if(x>=0&&x<n&&y>=0&&y<m) return 1; return 0; } int find(int k) { int i,j,x,y,di,dj; x=k/m; y=k%m; for(i=0;i<4;i++) { di=dir[i][0]+x; dj=dir[i][1]+y; if(judge(di,dj)&&!g[di][dj]) { j=di*m+dj; if(!mark[j]) { mark[j]=1; if(link[j]==-1||find(link[j])) { link[j]=k; return 1; } } } } return 0; } int main() { int u,v,k,i,j; while(scanf("%d%d%d",&n,&m,&k)!=-1) { memset(g,0,sizeof(g)); for(i=0;i<k;i++) { scanf("%d%d",&v,&u); u--;v--; g[u][v]=1; } if((n*m-k)&1) { printf("NO\n"); continue; } int ans=0; memset(link,-1,sizeof(link)); for(i=0;i<n;i++) { for(j=0;j<m;j++) { if((i+j)%2==0||g[i][j]) //(i+j)奇偶性!!! continue; memset(mark,0,sizeof(mark)); ans+=find(i*m+j); } } //printf("%d\n",ans); if(ans*2==n*m-k) printf("YES\n"); else printf("NO\n"); } return 0; }
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