HDU 4462（暴力枚举）

因为题目当中的k比较小k <= 10，所以可以直接枚举，题目里面由两个trick， 一个是如果每个点都可以放稻草人的话，那么答案是0, 另外一个就是如果可以放稻草人的点不用被照到。知道了这两个基本上暴力既可以ac了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 55;
struct Point {
    int x, y;
    int r;
}p[maxn];
bool vis[maxn][maxn];
int cnt;
bool vis2[maxn][maxn];
bool judge(int n, int k, int num)
{
    memset(vis, false, sizeof(vis));
    cnt = 0;
    for (int i = 0; i < k; i++)
        if ((1 << i) & num) 
            cnt++;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            if (vis2[i][j]) continue;
            for (int q = 0; q < k; q++)
            {
                if ((1 << q) & num)
                {
                    if (abs(p[q + 1].x - i) + abs(p[q + 1].y - j) <= p[q + 1].r)
                    {
                        vis[i][j] = true;
                        break;
                    }
                }
            }
            if (!vis[i][j]) return false;
        }
    }
    return true;
}
int main()
{
    int n, k;
    while (~scanf("%d", &n) && n)
    {
        memset(vis2, false, sizeof(vis2));
        scanf("%d", &k);
        for (int i = 1; i <= k; i++)
            scanf("%d %d", &p[i].x, &p[i].y);
        for (int i = 1; i <= k; i++)
            scanf("%d", &p[i].r);
        for (int i = 1; i <= k; i++)
            vis2[p[i].x][p[i].y] = true;
        bool flag = true;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
            {
                if (!vis2[i][j]) {
                    flag = false;
                    break;
                }

            }
        
        if (flag)
        {
            printf("0\n");
            continue;
        }
        int ans = 15000;
        int maxx = (1 << k);
        for (int i = 1; i < maxx; i++)
        {
            if (judge(n, k, i))
            {
                ans = min(ans, cnt);
            }
        }
        if (ans > 10)
            puts("-1");
        else
            printf("%d\n", ans);
    }
    return 0;
}