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直接对每个格子进行dfs结果除以2可以得到答案但是有大量重复的结果,不好输出答案.

可以只对横纵坐标相加是奇数的格子dfs....

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)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <vector>

using namespace std;

const int dir_x[4]={-1,1,0,0};
const int dir_y[4]={0,0,-1,1};

int mp[120][120];
int n,m,k;

bool used[120][120];
int linker[120][120];

bool dfs(int x,int y)
{
    for(int i=0;i<4;i++)
    {
        int X=x+dir_x[i];
        int Y=y+dir_y[i];
        if(mp[X][Y]==1) continue;
        if(X>n||X<1||Y>m||Y<1) continue;
        if(used[X][Y]) continue;
        used[X][Y]=true;
        if(linker[X][Y]==-1||dfs(linker[X][Y]/1000,linker[X][Y]%1000))
        {
            linker[X][Y]=x*1000+y;
            return true;
        }
    }
    return false;
}

int hungary()
{
    int ret=0;
    memset(linker,-1,sizeof(linker));
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if((i+j)&1||mp[i][j]==1) continue;
            memset(used,false,sizeof(used));
            if(dfs(i,j)) ret++;
        }
    }
    return ret;
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF&&n&&m)
    {
        scanf("%d",&k);
        memset(mp,0,sizeof(mp));
        for(int i=0;i<k;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            mp[a][b]=1;
        }
        printf("%d\n",hungary());
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                int ii=linker[i][j]/1000; int jj=linker[i][j]%1000;
                if(jj!=-1)
                printf("(%d,%d)--(%d,%d)\n",i,j,ii,jj);
            }
        }
    }
    return 0;
}

HDOJ 1507 Uncle Tom's Inherited Land*,布布扣,bubuko.com

HDOJ 1507 Uncle Tom's Inherited Land*

标签：des   style   class   code   java   http   

原文地址：http://blog.csdn.net/ck_boss/article/details/34570315