poj 3684 Physics Experiment 弹性碰撞

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
#include <map>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;

typedef struct
{
    int x, y;
}Node;

const int maxn = 210;
const int walk[][2] = {1, 0, 0, 1, -1, 0, 0, -1};

int n, m;
char a[maxn][maxn];
Node v[20];

bool judge(int s, int sum, int turnid)
{
    int vid = 0;
    set<P> check;
    for(int i = 0; i < sum; i++)    //确定非转向灯是否能合法放置 并记录他们所能照亮的房间
    {
        if(s & (1 << i) && i != turnid)
        {
            if(v[i].x - 1 >= 1 && v[i].x - 1 <= n && a[v[i].x - 1][v[i].y] != ‘.‘)
                return 0;
            if(v[i].y + 1 >= 1 && v[i].y + 1 <= m && a[v[i].x][v[i].y + 1] != ‘.‘)
                return 0;

            check.insert(P(v[i].x, v[i].y));
            if(v[i].x - 1 >= 1)
                check.insert(P(v[i].x - 1, v[i].y));
            if(v[i].y + 1 <= m)
                check.insert(P(v[i].x, v[i].y + 1));
        }
    }

    for(int i = 0; i < 4; i++)  //枚举转向灯的四个方向
    {
        int cnt = check.size();
        int x1 = v[turnid].x;
        int y1 = v[turnid].y;
        int x2 = x1 + walk[i][0];
        int y2 = y1 + walk[i][1];
        int x3 = x1 + walk[(i+1)%4][0];
        int y3 = y1 + walk[(i+1)%4][1];

        if(x2 >= 1 && x2 <= n && y2 >= 1 && y2 <= m && a[x2][y2] != ‘.‘)
            continue;
        if(x3 >= 1 && x3 <= n && y3 >= 1 && y3 <= m && a[x3][y3] != ‘.‘)
            continue;

        if(!check.count(P(x1, y1)))
            cnt++;
        if(x2 >= 1 && x2 <= n && y2 >= 1 && y2 <= m && !check.count(P(x2, y2)))
            cnt++;
        if(x3 >= 1 && x3 <= n && y3 >= 1 && y3 <= m && !check.count(P(x3, y3)))
            cnt++;

        if(cnt == sum)  //已照亮的房间 == 需要被照亮的房间
            return 1;
    }

    return 0;
}

int main()
{
    //freopen("in.txt", "r", stdin);

    while(scanf("%d%d", &n, &m) == 2 && !(n == 0 && m == 0))
    {
        for(int i = 1; i <= n; i++)
            scanf("%s", a[i]+1);

        int cnt = 0;    //记录总共有几个格子可以放灯
        for(int i = 1; i <= n; i++) //先把小于等于15个格子提出来
            for(int j = 1; j <= m; j++)
            {
                if(a[i][j] == ‘.‘)
                {
                    v[cnt].x = i;
                    v[cnt].y = j;
                    cnt++;
                }
            }

        if(cnt == 0)    //特判
        {
            printf("0\n");
            continue;
        }

        int ans = 20;

        for(int i = 1; i < (1 << cnt); i++)
        {
            int lcnt = 0;   //记录有几个灯
            int ltmp[20];   //记录灯对应的v[]的下标
            for(int j = 0; j < cnt; j++)
                if(i & (1 << j))
                    ltmp[lcnt++] = j;

            for(int j = 0; j < lcnt; j++)
                if(judge(i, cnt, ltmp[j]) && lcnt < ans)
                    ans = lcnt;
        }

        if(ans == 20)
            printf("-1\n");
        else
            printf("%d\n", ans);
    }

    return 0;
}