HDU 4499 Cannon （暴力求解）

题意：给定一个n*m个棋盘，放上一些棋子，问你最多能放几个炮（中国象棋中的炮）。

析：其实很简单，因为棋盘才是5*5最大，那么直接暴力就行，可以看成一行，很水，时间很短，才62ms。

代码如下：

#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e4 + 5;
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[30];
int ans;

bool isly(int x){
    int b[] = { x+1, x-1, x+m, x-m};
    int d[] = { x/m*m+m-1, x/m*m, n-1, 0};
    int c[] = { 1, -1, m, -m};
    
    for(int i = 0; i < 4; ++i){
        bool ok = false;
        for(int j = b[i]; (i & 1 ? j >= d[i] : j <= d[i]); j += c[i]){
            if(ok && a[j] != -1){
                if(a[j] == 1)  return false;
                else break;
            }
            else if(!ok && a[j] != -1) ok = true;
        }
    }
    return true;
}

void dfs(int x, int cnt){
    ans = max(ans, cnt);
    for(int i = x; i < n; ++i){
        if(a[i] != -1 || !isly(i))  continue;
        a[i] = 1;
        dfs(i+1, cnt+1);
        a[i] = -1;
    }
}

int main(){
    int c;
    while(scanf("%d %d %d", &n, &m, &c) == 3){
        n *= m;
        memset(a, -1, sizeof(a));
        for(int i = 0; i < c; ++i){
            int x, y;
            scanf("%d %d", &x, &y);
            a[x*m+y] = 0;
        }

        ans = 0;
        dfs(0, 0);
        printf("%d\n", ans);
    }
    return 0;
}