Chessboard
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 12800 | Accepted: 4000 |
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 2
2 1
3 3
Sample Output
YES
Hint
A possible solution for the
sample input.
Source
POJ
Monthly,charlescpp
和 hdu 1507类似，构无向图然后判断匹配数是否等于合法的格数。
心算32*32错了= = RE了两次，开始以为32*32是90+，第二次以为是900+，笔算后才知道是1024..
1 //224K 125MS C++ 1731B 2014-06-10 12:44:41
2 #include<iostream>
3 #include<vector>
4 #define N 1050
5 using namespace std;
6 vector<int>V[N];
7 int match[N];
8 int vis[N];
9 int g[35][35];
10 int dfs(int u)
11 {
12 for(int i=0;i<V[u].size();i++){
13 int v=V[u][i];
14 if(!vis[v]){
15 vis[v]=1;
16 if(match[v]==-1 || dfs(match[v])){
17 match[v]=u;
18 return 1;
19 }
20 }
21 }
22 return 0;
23 }
24 int hungary(int n)
25 {
26 int ret=0;
27 memset(match,-1,sizeof(match));
28 for(int i=1;i<=n;i++){
29 memset(vis,0,sizeof(vis));
30 ret+=dfs(i);
31 }
32 return ret;
33 }
34 int main(void)
35 {
36 int n,m,k,x,y;
37 while(scanf("%d%d%d",&n,&m,&k)!=EOF)
38 {
39 memset(g,0,sizeof(g));
40 for(int i=0;i<N;i++) V[i].clear();
41 for(int i=0;i<k;i++){
42 scanf("%d%d",&y,&x);
43 g[x-1][y]=1;
44 }
45 int map[N]={0},pos=0;
46 for(int i=0;i<n;i++)
47 for(int j=1;j<=m;j++)
48 if(!g[i][j]){
49 if(!map[i*m+j]) map[i*m+j]=++pos;
50 int u=map[i*m+j];
51 if(j<m && !g[i][j+1]){
52 if(!map[i*m+j+1]) map[i*m+j+1]=++pos;
53 V[u].push_back(map[i*m+j+1]);
54 V[map[i*m+j+1]].push_back(u);
55 }
56 if(i<n-1 && !g[i+1][j]){
57 if(!map[(i+1)*m+j]) map[(i+1)*m+j]=++pos;
58 V[u].push_back(map[(i+1)*m+j]);
59 V[map[(i+1)*m+j]].push_back(u);
60 }
61 }
62 //printf("%d\n",pos);
63 if(hungary(pos)==pos) puts("YES");
64 else puts("NO");
65 }
66 return 0;
67 }