hdu 4770 Lights Against Dudely(回溯)

pid=4770" target="_blank" style="">题目链接：hdu 4770 Lights Against Dudely

题目大意：在一个N*M的银行里。有N*M个房间，‘#’代表牢固的房间，‘.‘代表的是脆弱的房间。脆弱的房间个数不会超过15个，如今为了确保安全，要在若干个脆弱的房间上装灯。普通的灯是照亮{0, 0}, {-1, 0}, {0, 1}(和题目中坐标有点出入)。然后能够装一个特殊的，能够照耀

• { {0, 0}, {0, 1}, {1, 0} },
• { {0, 0}, {-1, 0}, {0, -1} },
• { {0, 0}, {0, -1}, {1, 0} }
  
  同一个房间不能够装两栈灯，灯光不能照耀牢固的房间，问说最少须要多少栈灯。

解题思路：dfs+剪枝。暴力枚举放特殊灯的位置，然后将脆弱房间依照i坐标大放前面，相等的将j坐标小的方前面，这样做是为了dfs的时候剪枝，仅仅要碰到一个房间不能放就返回。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 200;
const int maxv = 20;
const int INF = 0x3f3f3f3f;
const int dir[4][3][2] = { { {0, 0}, {-1, 0}, {0, 1} },
                           { {0, 0}, {0, 1}, {1, 0} },
                           { {0, 0}, {-1, 0}, {0, -1} },
                           { {0, 0}, {0, -1}, {1, 0} }
                        };

int ans;
int n, N, M, x[maxv], y[maxv], c[maxv];
int v[maxn+5][maxn+5];
char g[maxn+5][maxn+5];

inline int judge (int xi, int yi, const int d[3][2]) {

    for (int i = 0; i < 3; i++) {
        int p = xi + d[i][0];
        int q = yi + d[i][1];

        if (p <= 0 || p > N)
            continue;

        if (q <= 0 || q > M)
            continue;

        if (g[p][q] == '#')
            return 0;
    }
    return 1;
}

inline void set (int xi, int yi, const int d[3][2], int type) {
    for (int i = 0; i < 3; i++) {
        int p = xi + d[i][0];
        int q = yi + d[i][1];

        if (p <= 0 || p > N)
            continue;

        if (q <= 0 || q > M)
            continue;

        v[p][q] = type;
    }
}

void init () {
    n = 0;
    for (int i = 1; i <= N; i++)
        scanf("%s", g[i] + 1);

    for (int i = N; i; i--) {
        for (int j = 1; j <= M; j++) {
            if (g[i][j] == '.') {
                x[n] = i;
                y[n] = j;
                n++;
            }
        }
    }

    memset(c, 0, sizeof(c));
    for (int i = 0; i < n; i++)
        c[i] = judge(x[i], y[i], dir[0]);
}

/*
   int solve (int spi, int id) {

   memset(v, 0, sizeof(v));
   int ans = INF;
   for (int s = 0; s < (1<<n); s++) {

   bool flag = true;
   for (int i = 0; i < n; i++) {
   if (s&(1<<i) && (c[i] == 0 || i == spi)) {
   flag = false;
   break;
   }
   }

   if (flag) {
   int light = 0;
   int tmp = set(x[spi], y[spi], dir[id], 1);

   for (int i = 0; i < n; i++) {
   if (s&(1<<i)) {
   light++;
   tmp += set(x[i], y[i], dir[0], 1);
   }
   }

   if (tmp == n)
   ans = min(ans, light);

   memset(v, 0, sizeof(v));
   }
   }
   return ans+1;
   }
   */

void dfs (int d, int f, int cnt) {

    if (cnt >= ans)
        return;

    if (d == n) {
        ans = cnt;
        return;
    }

    if (v[x[d]][y[d]])
        dfs (d + 1, f, cnt);

    if (c[d] && d != f) {
        set(x[d], y[d], dir[0], 1);
        dfs (d + 1, f, cnt+1);
        set(x[d], y[d], dir[0], 0);
    }
}

int main () {
    while (scanf("%d%d", &N, &M) == 2 && N + M) {
        init();

        ans = INF;
        if (n == 0)
            ans = 0;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 4; j++) {
                if (judge (x[i], y[i], dir[j])) {
                    memset(v, 0, sizeof(v));

                    set(x[i], y[i], dir[j], 1);

                    dfs(0, i, 1);

                    set(x[i], y[i], dir[j], 0);
                }
            }
        }

        if (ans == INF)
            printf("-1\n");
        else
            printf("%d\n", ans);
    }
    return 0;
}