CodeFroces Round 423 B. Black Square(暴力枚举)

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http://codeforces.com/contest/828/problem/B

题目大意，给出一个n*m的纸，里面的方格为白色或者黑色，要求描最少的白色格子使得黑色为一个正方形。如果不存在这种解法则输出-1.

解法：先得到黑色格子的最上最下最左最右的坐标，然后枚举。

代码如下：

#include<bits/stdc++.h>using namespace std;char G[105][105];int find(int lx, int rx, int ty, int ly) {int cnt = 0;for(int j = ty; j <= ly; j++) {for(int i = lx; i <= rx; i++) {if(G[j][i] != 'B') {cnt++;}}}return cnt;}int main() {int n, m, lx = 105, rx = -1, ty = 105, ly = -1, ans = 0x3f3f3f3f, big;cin >> n >> m;for(int i = 0; i < n; i++)scanf("%s", &G[i]);for(int i = 0; i < n; i++) {for(int j = 0; j < m; j++) {if(G[i][j] == 'B') {if(j < lx)lx = j;if(j > rx)rx = j;if(i < ty)ty = i;if(i > ly)ly = i;}}}if(lx != 105 && rx - lx >= ly - ty) {big = rx - lx;for(int i = 0; i + big < n; i++) {if(i <= ty && i + big >= ty) {ans = min(ans, find(lx, rx, i, i + big));}}} if(lx != 105 && rx - lx <= ly - ty) {big = ly - ty;for(int i = 0; i + big < m; i++) {if(i <= lx && i + big >= rx) {ans = min(ans, find(i, i + big, ty, ly));}}}if(lx != 105 && ans == 0x3f3f3f3f)cout << "-1" << endl;else if(ans == 0x3f3f3f3f) cout << "1" << endl;elsecout << ans << endl;return 0;}

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