Description

Thanks to a certain "green" resources company, there is a new profitable industry of oil skimming. There are large slicks of crude oil floating in the Gulf of Mexico just waiting to be scooped up by enterprising oil barons. One such oil baron has a special plane that can skim the surface of the water collecting oil on the water‘s surface. However, each scoop covers a 10m by 20m rectangle (going either east/west or north/south). It also requires that the rectangle be completely covered in oil, otherwise the product is contaminated by pure ocean water and thus unprofitable!

Given a map of an oil slick, the oil baron would like you to compute the maximum number of scoops that may be extracted. The map is an NxN grid where each cell represents a 10m square of water, and each cell is marked as either being covered in oil or pure water.

Input

Output

Sample Input

16.......##....##.......#.....##......

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#include <vector>using namespace std;#define N 1210int cx[N];int cy[N];int nx,ny;int mk[N];vector<int> map[N];int ma[N][N];char g[610][610];int path(int u){    int len = map[u].size();    for(int i = 0; i < len; i ++)    {        int v = map[u][i];        if(!mk[v])        {            mk[v] = 1;            if(cy[v] == -1 || path(cy[v]))  ///cy #号块也没有动||cy 也有符合条件的            {                cx[u] = v;                cy[v] = u;                return 1;            }        }    }    return 0;}int maxma(){    int res = 0;    memset(cx,-1,sizeof(cx));    memset(cy,-1,sizeof(cy));    for(int i = 0; i < nx; i ++)    {        if(cx[i] == -1)   ///#号块儿 没动        {            memset(mk,0,sizeof(mk));            res += path(i);            //printf("%d---\n",res);        }    }    return res;}int main(){    int t,n;memset(g,0,sizeof(g));    //freopen("a.txt","r",stdin);    scanf("%d",&t);    int ca = 1;    while(t--)    {        scanf("%d",&n);        for(int i = 1;i <= n*n;i ++)           map[i].clear();  ///初始化        int num = 0;        //memset(map,0,sizeof(map));        for(int i = 1; i <= n; i ++)        {            scanf("%s",g[i]+1);            for(int j = 1;j <= n;j ++)               if(g[i][j]==‘#‘) ma[i][j] = num++;  ///多少#            //printf("||%s\n",g[i]+1);        }        for(int i = 1; i <= n; i ++)            for(int j = 1; j <= n; j ++)   ///符合条件的            {                if(g[i][j] != ‘#‘) continue;                if(g[i][j] == ‘#‘ && ‘#‘ == g[i+1][j])                    map[ma[i][j]].push_back(ma[i+1][j]);                if(g[i][j] == ‘#‘ && g[i-1][j] == ‘#‘)                    map[ma[i][j]].push_back(ma[i-1][j]);                if(g[i][j] == ‘#‘ && g[i][j+1] == ‘#‘)                    map[ma[i][j]].push_back(ma[i][j+1]);                if(g[i][j] == ‘#‘ && g[i][j-1] == ‘#‘)                    map[ma[i][j]].push_back(ma[i][j-1]);            }        nx = ny = num;        printf("Case %d: %d\n",ca++,maxma()/2);    }    return 0;}