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意甲冠军：特定n*m矩阵

X代表色 .代表无色

随着x*y形刷子去涂色。

刷子每次能够→或↓移动随意步。

若可以染出给定的矩阵，则输出最小的刷子的面积

若不能输出-1

思路：

先找到连续最小的x,y

由于至少一个边界和x或y相等，所以枚举(x,i) 和 (i,y)就能够了。

#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
using namespace std;
#define N 1010
int top;
char s[N][N];
int n, m, stx, sty;
int H(int h, int l, int r){//扫一行 -1表示全是. 1表示全是X 0表示都有if(h>n || r>m) return -2;int cnt = 0;for(int i = l; i <= r; i++)if(s[h][i]=='X')cnt++;if(cnt==0)return -1;if(cnt==r-l+1)return 1;return 0;
}
int L(int l, int S, int X){if(l>m || X>n) return -2;int cnt = 0;for(int i = S; i <= X; i++)if(s[i][l]=='X')cnt++;if(cnt==0)return -1;if(cnt==X-S+1)return 1;return 0;
}
bool ok(int x, int y){//  printf(" ++++++(%d,%d)\n", x,y);int nowx = stx, nowy = sty;if(nowx + x-1 > n || nowy + y-1>m)return false;for(int i = 0; i < x; i++)for(int j = 0; j < y; j++)if(s[i+nowx][j+nowy]!='X')return false;//   puts("---");put(); puts("-----");int cnt = x*y;while(1){if(nowx + x <= n && s[nowx+x][nowy] =='X'){nowx ++;for(int i = 0; i < y; i++)if(s[nowx+x-1][i+nowy]!='X')return false;cnt += y;}else if(nowy + y <= m && s[nowx][nowy+y] == 'X'){nowy++;for(int i = 0; i < x; i++)if(s[i+nowx][nowy+y-1]!='X')return false;cnt += x;}else break;//    puts("******");put();}return cnt == top;
}
int hehe; int feifei;
int x, y;
void find_xy(){x = N, y = N;for(int i = 1; i <= m; i++) {int cnt = 0;for(int j = 1; j <= n; j++){if(s[j][i]=='.'){if(cnt) x = min(x, cnt);cnt = 0;}elsecnt++;}if(cnt) x = min(x, cnt);}for(int i = 1; i <= n; i++) {int cnt = 0;for(int j = 1; j <= m; j++){if(s[i][j]=='.'){if(cnt) y = min(y, cnt);cnt = 0;}else cnt++;}if(cnt) y = min(y, cnt);}
}
int solve(){int ans = N*N;for(int i = 1; i <= x; i++)if(ok(i,y)) {ans = i*y;break;}for(int i = 1; i <= y && x*i<ans; i++)if(ok(x,i)){ans = x*i;break;}if(ans > n*m) return -1;return ans;
}
void input(){stx = -1;top = 0;for(int i = 1; i <= n; i++){scanf("%s", s[i]+1);for(int j = 1; j <= m; j++) {if(stx==-1 && s[i][j]=='X'){stx = i, sty = j;}top += s[i][j]=='X';}}
}
int main(){while(cin>>n>>m){input();find_xy();printf("%d\n", solve());}return 0;
}