标签：二分匹配

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)

二分匹配，建边的依据是枚举所有可行的点，然后枚举周围四个方向；输出方案依据mark数组

#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;
const int maxn = 10010;  
struct node  
{  
    int to;  
    int next;  
}edge[4 * maxn];  

int vis[maxn];
int head[maxn];  
int mark[maxn];  
bool used[maxn];  
bool mat[110][110];  
int cnt[110][110];  
int tot;  
int n, m;  
int index;  
  
void addedge(int from, int to)  
{  
    edge[tot].to = to;  
    edge[tot].next = head[from];  
    head[from] = tot++;   
}  
  
bool dfs(int x)  
{  
    for (int i = head[x]; i != -1; i = edge[i].next)  
    {  
        if (!used[edge[i].to])  
        {  
            used[edge[i].to] = 1;  
            if (mark[edge[i].to] == -1 || dfs(mark[edge[i].to]))  
            {  
                mark[edge[i].to] = x;
                return true;  
            }  
        }  
    }  
    return false;  
}  
  
int hungary()  
{  
    memset(mark, -1, sizeof(mark));  
    int ans = 0;  
    for (int i = 0; i < index; i++)  
    {  
        memset(used, 0, sizeof(used));  
        if (dfs(i))  
            ans++;  
    }  
    return ans;  
}  
  
int main()  
{  
    int k;  
    while (~scanf("%d%d", &n, &m))  
    {  
    	if (!n && !m)
    	{
    		break;
    	}
    	scanf("%d", &k);
    	map<int, int>qu;
    	qu.clear();
        memset(head, -1, sizeof(head));  
        memset(mat, false, sizeof(mat)); 
        memset(vis, -1, sizeof(vis));
        tot=0;  
        int x, y;  
        for (int i = 0; i < k; i++)  
        {  
            scanf("%d%d", &x, &y);  
            x--;  
            y--; 
            mat[x][y] = true;  
        }  
        index = 0;  
        for (int i = 0; i < n; i++)  
            for (int j = 0; j < m; j++)  
                if (!mat[i][j])  
                {
                	cnt[i][j] = index++; 
                	qu[index - 1] = i * m + j;
                }
                     
        for (int i = 0; i < n; i++)  
            for (int j = 0; j < m; j++)  
            {   
                if (!mat[i][j])  
                {  
                    if (i > 0 && !mat[i - 1][j])  
                        addedge(cnt[i][j], cnt[i - 1][j]);  
                    if (i < n-1 && !mat[i + 1][j])  
                        addedge(cnt[i][j], cnt[i + 1][j]);  
                    if (j > 0 && !mat[i][j - 1])  
                        addedge(cnt[i][j], cnt[i][j - 1]);  
                    if (j < m - 1 && !mat[i][j + 1])  
                        addedge(cnt[i][j], cnt[i][j + 1]);  
                }  
            }  
        int res = hungary();  
        printf("%d\n", res / 2);
        for (int i = 0; i < index; ++i)
        {
        	if (mark[i] != -1)
        	{
        		int a = qu[i];
        		int b = qu[mark[i]];
        		int x1 = a / m + 1;
        		int y1 = a % m + 1;
        		int x2 = b / m + 1;
        		int y2 = b % m + 1;
        		if (vis[a] == -1 && vis[b] == -1)
        		{
        			printf("(%d,%d)--(%d,%d)\n", x1, y1, x2, y2);
        			vis[a] = b;
        			vis[b] = a;
        		}
        	}
        }
        printf("\n");
    }  
    return 0;  
} 

hdu1507——Uncle Tom's Inherited Land*

标签：二分匹配

原文地址：http://blog.csdn.net/guard_mine/article/details/41315879