POJ 2226 Muddy Fields

Muddy Fields

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12683		Accepted: 4696

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

4

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

 

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
//#include<queue>
//#include<set>
#include<bitset>
//#include<cassert>
#include<vector>
#define INF 0x3f3f3f3f
#define re register
#define Ii inline int
#define Il inline long long
#define Iv inline void
#define Ib inline bool
#define Id inline double
#define ll long long
#define Fill(a,b) memset(a,b,sizeof(a))
#define R(a,b,c) for(register int a=b;a<=c;++a)
#define nR(a,b,c) for(register int a=b;a>=c;--a)
#define Min(a,b) ((a)<(b)?(a):(b))
#define Max(a,b) ((a)>(b)?(a):(b))
#define Cmin(a,b) ((a)=(a)<(b)?(a):(b))
#define Cmax(a,b) ((a)=(a)>(b)?(a):(b))
#define abs(a) ((a)>0?(a):-(a))
#define D_e(x) printf("&__ %d __&
",x)
#define D_e_Line printf("-----------------
")
#define Pause system("pause");
using namespace std;
const int N=2505;
Ii read(){
    int s=0,f=1;char c;
    for(c=getchar();c>'9'||c<'0';c=getchar())if(c=='-')f=-1;
    while(c>='0'&&c<='9')s=s*10+(c^'0'),c=getchar();
    return s*f;
}
Iv print(int x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)print(x/10);
    putchar(x%10^'0');
}
using namespace std;
int vis[N],match[N],tim;
vector<int>V[N];
Ib Hungary(int u){
    for(vector<int>::iterator v=V[u].begin();v!=V[u].end();++v)
        if(vis[*v]!=tim){
            vis[*v]=tim;
            if(!match[*v]||Hungary(match[*v])){
                match[*v]=u;return 1;
            }
        }
    return 0;
}
char mp[55][55];
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        R(i,0,n)
            vector<int>().swap(V[i]),match[i]=0,vis[i]=0;
        char *s=new char[10];
        R(i,1,n){
            scanf("%s",s+1);
            R(j,1,m)
                mp[i][j]=s[j];
        }
        delete []s;
        int **a=new int *[n+1];
        int **b=new int *[n+1];
        R(i,0,n+1){
            a[i]=new int[m+1];
            b[i]=new int[m+1];
        }
        Fill(a,0),Fill(b,0);
        int x=0,y=0;
        R(i,1,n)
            R(j,1,m)
                if(mp[i][j]=='*')
                    a[i][j]=(mp[i][j-1]=='*')?a[i][j-1]:++x;
        R(i,1,n)
            R(j,1,m)
                if(mp[i][j]=='*'){
                    b[i][j]=(mp[i-1][j]=='*')?b[i-1][j]:++y;
                    V[a[i][j]].push_back(b[i][j]);
                }
        R(i,0,n+1){
            delete []a[i];
            delete []b[i];    
        }
        delete []a;
        delete []b;
        int ans=0;
        tim=0;
        R(i,1,x)
            ++tim,
            ans+=Hungary(i);
        print(ans),putchar('
');
    }
    return 0;    
}