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Uncle Tom‘s Inherited Land*

Your old uncle Tom inherited a piece of land from his great-great-uncle. Originally, the property had been in the shape of a rectangle. A long time ago, however, his great-great-uncle decided to divide the land into a grid of small squares. He turned some of the squares into ponds, for he loved to hunt ducks and wanted to attract them to his property. (You cannot be sure, for you have not been to the place, but he may have made so many ponds that the land may now consist of several disconnected islands.)

Your uncle Tom wants to sell the inherited land, but local rules now regulate property sales. Your uncle has been informed that, at his great-great-uncle‘s request, a law has been passed which establishes that property can only be sold in rectangular lots the size of two squares of your uncle‘s property. Furthermore, ponds are not salable property.

Your uncle asked your help to determine the largest number of properties he could sell (the remaining squares will become recreational parks). 

4 4
6
1 1
1 4
2 2
4 1
4 2
4 4
4 3
4
4 2
3 2
2 2
3 1
0 0

4
(1,2)--(1,3)
(2,1)--(3,1)
(2,3)--(3,3)
(2,4)--(3,4)

3
(1,1)--(2,1)
(1,2)--(1,3)
(2,3)--(3,3)

思路：
编号，构建新图，遍历坐标和为奇数的点，求出最大匹配；
然后，遍历每个点（link数组），如果它有匹配点的话，就将其输出。

#include"stdio.h"
#include"string.h"
#define N 100
#define M 55
int e[N][N],id[N][N];
int n,m,g[N][N],cnt;
int mark[N],link[N];
int dir[4][2]={0,1,0,-1,-1,0,1,0};
struct node         //存可匹配点的横纵坐标
{
    int x,y;
}p[N];
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;
    for(i=0;i<cnt;i++)
    {
        if((p[i].x+p[i].y)%2)
            continue;
        if(!mark[i]&&g[k][i])
        {
            mark[i]=1;
            if(link[i]==-1||find(link[i]))
            {
                link[i]=k;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,j,k,q,x,y,u,v;
    while(scanf("%d%d",&n,&m),n||m)
    {
        memset(e,-1,sizeof(e));
        memset(g,0,sizeof(g));
        memset(id,0,sizeof(id));
        scanf("%d",&q);
        while(q--)
        {
            scanf("%d%d",&u,&v);
            e[u][v]=0;
        }
        cnt=0;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(e[i][j]==-1)
                {
                    p[cnt].x=i;
                    p[cnt].y=j;
                    id[i][j]=cnt++;
                }
            }
        }
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                if(e[i][j]==-1)
                {
                    for(k=0;k<4;k++)
                    {
                        x=dir[k][0]+i;
                        y=dir[k][1]+j;
                        if(judge(x,y)&&e[x][y]==-1)
                        {
                            u=id[i][j];
                            v=id[x][y];
                            g[u][v]=g[v][u]=1;
                        }
                    }
                }
            }
        }
        int ans=0;
        memset(link,-1,sizeof(link));
        for(i=0;i<cnt;i++)
        {
            if((p[i].x+p[i].y)%2==0)
                continue;
            memset(mark,0,sizeof(mark));
            ans+=find(i);
        }
        printf("%d\n",ans);
        for(i=0;i<cnt;i++)
        {
            if(link[i]==-1)
                continue;
            j=link[i];
            printf("(%d,%d)--(%d,%d)\n",p[i].x,p[i].y,p[j].x,p[j].y);
        }
        printf("\n");
    }
    return 0;
}

hdu 1507 Uncle Tom's Inherited Land*（二分）,布布扣,bubuko.com

hdu 1507 Uncle Tom's Inherited Land*（二分）

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原文地址：http://blog.csdn.net/u011721440/article/details/38144629