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;
}
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