[面试备] 暴搜 or 二分图的经典升级 ： hdu 1045 Fire Net 示例 [ 讲解之用 ]

题意：
    现有一座n*n（0<=n<=4)的大城市，要你在没有障碍的地方放炮台，每个炮塔都会有向四面射击的机枪，子弹能够击穿本行本列所有除障碍物的其他物品。问你最多能放多少炮台。
解法：
    dfs+回溯 如果题目数据大一点还要用到二分匹配。我好喜欢的一道题目！

#include <iostream>#include <string>#include <cstring>#include <cstdlib>#include <cstdio>#include <cmath>#include <vector>#include <stack>#include <deque>#include <queue>#include <bitset>#include <list>#include <map>#include <set>#include <iterator>#include <algorithm>#include <functional>#include <utility>#include <sstream>#include <climits>#include <cassert>#define BUG puts("here!!!");using namespace std;const int N = 4;char mmap[N][N];int n, mmax;bool ok(int x, int y) {if(mmap[x][y] != '.') return false;        for(int i = y-1; i >= 0; i--) {if(mmap[x][i] == 'X') break;if(mmap[x][i] == 'B') return false;}for(int i = x-1; i >= 0; i--) {if(mmap[i][y] == 'X') break;if(mmap[i][y] == 'B') return false;}return true;}void dfs(int pos, int num) {if(pos == n*n) {if(num > mmax) mmax = num;return;}int x = pos / n;int y = pos % n;if(ok(x, y)) {mmap[x][y] = 'B';dfs(pos+1, num+1);mmap[x][y] = '.';}dfs(pos+1, num);}int main() {while(cin >> n, n) {mmax = -1;for(int i = 0; i < n; i++) {for(int j = 0; j < n; j++) {cin >> mmap[i][j];}}dfs(0, 0);cout << mmax << endl;}return 0;}