2226 Muddy Fields //MaxMatch 最大点集覆盖的变形

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本文探讨了一种算法挑战，即如何使用最少数量的木板覆盖一个矩形田野中的泥泞区域，确保奶牛可以在干净的草地上吃草。田野由行和列组成，每个泥泞斑块可通过横向或纵向放置的木板覆盖。目标是最小化所需木板的数量。

Muddy Fields

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3483		Accepted: 1299

Description

Rain has pummeled the cows' field, a rectangular grid of R rows and C columns (1 <= R <= 50, 1 <= C <= 50). While good for the grass, the rain makes some patches of bare earth quite muddy. The cows, being meticulous grazers, don't want to get their hooves dirty while they eat.

To prevent those muddy hooves, Farmer John will place a number of wooden boards over the muddy parts of the cows' field. Each of the boards is 1 unit wide, and can be any length long. Each board must be aligned parallel to one of the sides of the field.

Farmer John wishes to minimize the number of boards needed to cover the muddy spots, some of which might require more than one board to cover. The boards may not cover any grass and deprive the cows of grazing area but they can overlap each other.

Compute the minimum number of boards FJ requires to cover all the mud in the field.

Input

* Line 1: Two space-separated integers: R and C

* Lines 2..R+1: Each line contains a string of C characters, with '*' representing a muddy patch, and '.' representing a grassy patch. No spaces are present.

Output

* Line 1: A single integer representing the number of boards FJ needs.

Sample Input

4 4
*.*.
.***
***.
..*.

Sample Output

Hint

OUTPUT DETAILS:

Boards 1, 2, 3 and 4 are placed as follows:
1.2.
.333
444.
..2.
Board 2 overlaps boards 3 and 4.

Source

USACO 2005 January Gold

把行里面连在一起的坑连起来视为一个点，即一块横木板，编上序号，Sample则转化为：

1 0 2 0
0 3 3 3
4 4 4 0
0 0 5 0

把这些序号加入X集合，再按列做一次则为：

1 0 4 0
0 3 4 5
2 3 4 0
0 0 4 0

同样加入Y集合，一个坑只能被横着的或者被竖着的木板盖住，将原图的坑的也标上不同的序号，一共九个坑

1 . 2 .
. 3 4 5
67 8 .
. . 9 .

比如7号坑可以被横着的4号木板和竖着的3号木板盖住，把每个点的对应的横木板(4)和竖木板(3)中间连一条边的话,则问题转化为找尽量少的边把这些点都盖住,根据xxxxx定理便是求最大匹配数

#include<stdio.h>
#include<string.h>
int r,c;
char str[51][51];
int x[51][51],y[51][51];
int mat[1255][1255];
int link[1255];
bool usedif[1255];
int numb,numa;
bool can(int t)
{
    for(int i=1;i<=numb;i++)
    if(usedif[i]==0&&mat[t][i]==1)
    {
        usedif[i]=1;
        if(link[i]==-1||can(link[i]))
        {
            link[i]=t;
            return true;
        }
    }
    return false;
}
int MaxMatch()
{
    int num=0;
    memset(link,-1,sizeof(link));
    for(int i=1;i<=numa;i++)
    {
        memset(usedif,0,sizeof(usedif));
        if(can(i)) num++;
    }
    return num;
}
int main()
{
    while(scanf("%d%d",&r,&c)!=EOF)
    {
        for(int i=0;i<r;i++) scanf("%s",str[i]);
        numa=0;
        for(int i=0;i<r;i++)
          for(int j=0;j<c;j++)
          {
              if(j==0)
              {
                  if(str[i][j]=='*') x[i][j]=++numa;
              }
              else
              {
                  if(str[i][j]=='*')
                  {
                      if(str[i][j]==str[i][j-1]) x[i][j]=numa;
                      else x[i][j]=++numa;
                  }
              }
          }
        numb=0;
        for(int j=0;j<c;j++)
          for(int i=0;i<r;i++)
          {
              if(i==0)
              {
                  if(str[i][j]=='*') y[i][j]=++numb;
              }
              else
              {
                  if(str[i][j]=='*')
                  {
                      if(str[i][j]==str[i-1][j]) y[i][j]=numb;
                      else y[i][j]=++numb;
                  }
              }
          }
        memset(mat,0,sizeof(mat));
        for(int i=0;i<r;i++)
          for(int j=0;j<c;j++)
          {
              mat[x[i][j]][y[i][j]]=1;
          }
        printf("%d/n",MaxMatch());
    }
    return 0;
}