Chessboard

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11078		Accepted: 3449

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

,charlescpp

 

 

 

独立集：

    独立集是指图的顶点集的一个子集,该子集的导出子图不含边.如果一个独立集不是任何一个独立集的子集, 那么称这个独立集是一个极大独立集.一个图中包含顶点数目最多的独立集称为最大独立集。最大独立集 一定是极大独立集，但是极大独立集不一定是最大的独立集。

支配集：

    与独立集相对应的就是支配集，支配集也是图顶点集的一个子集，设S 是图G 的一个支配集，则对于图中的任意一个顶点u，要么属于集合s, 要么与s 中的顶点相邻。 在s中除去任何元素后s不再是支配集，则支配集s是极小支配集。称G的所有支配集中顶点个数最 少的支配集为最小支配集，最小支配集中的顶点个数成为支配数。

最小点的覆盖：

    最小点的覆盖也是图的顶点集的一个子集，如果我们选中一个点，则称这个点将以他为端点的所有边都覆盖了。将图中所有的边都覆盖所用顶点数最少，这个集合就是最小的点的覆盖。

最大团：

    图G的顶点的子集，设D是最大团，则D中任意两点相邻。若u，v是最大团，则u,v有边相连，其补图u,v没有边相连，所以图G的最大团=其补图的最大独立集。

一些性质：

最大独立集+最小覆盖集=V

最大团=补图的最大独立集

最小覆盖集=最大匹配

 

#include
      
       #include
       
        #include
        
         using namespace std;int m,n,k,tot;int map[1610][1610],g[40][40],tg[40][40];int linker[1610],vis[1610];int DFS(int u){    int v;    for(v=1;v<=tot;v++)        if(map[u][v] && !vis[v]){            vis[v]=1;            if(linker[v]==-1 || DFS(linker[v])){                linker[v]=u;                return 1;            }        }    return 0;}int Hungary(){    int u,ans=0;    memset(linker,-1,sizeof(linker));    for(u=1;u<=tot;u++){        memset(vis,0,sizeof(vis));        if(DFS(u))            ans++;    }    return ans;}int main(){    //freopen("input.txt","r",stdin);    while(~scanf("%d%d%d",&m,&n,&k)){        int x,y;        memset(g,0,sizeof(g));        while(k--){            scanf("%d%d",&x,&y);            g[y][x]=1;   //注意这里是y行x列，被WA了好几次。。。。。。。。。。。。        }        tot=0;        memset(tg,0,sizeof(tg));        for(int i=1;i<=m;i++)            for(int j=1;j<=n;j++)                if(g[i][j]==0)                    tg[i][j]=++tot;        memset(map,0,sizeof(map));        for(int i=1;i<=m;i++)            for(int j=1;j<=n;j++)                if(tg[i][j]!=0){                    if(i>=1 && tg[i-1][j]!=0)                        map[tg[i][j]][tg[i-1][j]]=1;                    if(i<=m && tg[i+1][j]!=0)                        map[tg[i][j]][tg[i+1][j]]=1;                    if(j>=1 && tg[i][j-1]!=0)                        map[tg[i][j]][tg[i][j-1]]=1;                    if(j<=n && tg[i][j+1]!=0)                        map[tg[i][j]][tg[i][j+1]]=1;                }        int ans=Hungary();        if(ans==tot)            puts("YES");        else            puts("NO");    }    return 0;}