HDU

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

The input starts with an integer K (1 <= K <= 100) indicating the number of cases. Each case starts with an integer N (1 <= N <= 600) indicating the size of the square grid. Each of the following N lines contains N characters that represent the cells of a row in the grid. A character of '#' represents an oily cell, and a character of '.' represents a pure water cell.

Output

For each case, one line should be produced, formatted exactly as follows: "Case X: M" where X is the case number (starting from 1) and M is the maximum number of scoops of oil that may be extracted.

Sample Input

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

Sample Output

Case 1: 3

这道题目是说给你一个规模为n的方阵，里面的‘#’号是指油田，'.'是指水。当有10*20大小的矩形存在的时候，表示才能够提取出油来。

问所给的矩阵中符合条件的情况中最大能提取出多少油来。

思路：二分匹配+dfs。先遍历所有的字符。如果是‘#’，就标记坐标和个数，，记录个数最后直接dfs，降低复杂度。

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <vector>#include <cmath>#include <algorithm>using namespace std;#define MAX_N 100005#define INF 0x3f3f3f3f#define Mem(a,x) memset(a,x,sizeof(a))#define ll long longchar m[605][605];int mp[605][605],gra[605][605],link[10005];bool used[360005];int N;bool dfs(int u) {    for(int v = 0; v<N; v++) {        if(!gra[u][v] || used[v]) continue;        used[v] = true;        if(link[v] == -1 || dfs(link[v])) {            link[v] = u;            return true;        }    }    return false;}int solve() {    Mem(link,-1);    int sum = 0;    for(int i = 0; i<N; i++) {        Mem(used,false);        used[i] = true;        if(dfs(i)) sum ++;    }    return sum;}int main(){    int k;    cin>>k;    int num = 1;    while(k--) {        int n;        cin>>n;        N = 0;        Mem(mp,0);        Mem(gra,0);        for(int i = 0; i<n; i++) {            scanf("%s",&m[i]);        }        for(int i = 0; i<n; i++) {            for(int j = 0; j<n; j++) {                if(m[i][j] == '#') {                    mp[i][j] = N++;                }            }        }        for(int i = 0; i<n; i++) {            for(int j = 0; j<n; j++) {                if(m[i][j] == '.') continue;                if(i-1>=0 && m[i-1][j] == '#') gra[mp[i][j]][mp[i-1][j]] = 1;                if(i+1<n && m[i+1][j] == '#') gra[mp[i][j]][mp[i+1][j]] = 1;                if(j-1>=0 && m[i][j-1] == '#') gra[mp[i][j]][mp[i][j-1]] = 1;                if(j+1<n && m[i][j+1] == '#') gra[mp[i][j]][mp[i][j+1]] = 1;            }        }        int ans = solve() / 2;        printf("Case %d: %d\n",num ++,ans);    }    return 0;}