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,‘#’代表坚固的房间,‘.‘代表的是脆弱的房间,脆弱的房间个数不会超过15个,现在为了确保安全,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),然后可以装一个特殊的,可以照射
- { {0, 0}, {0, 1}, {1, 0} },
- { {0, 0}, {-1, 0}, {0, -1} },
- { {0, 0}, {0, -1}, {1, 0} }
同一个房间不可以装两栈灯,灯光不能照射坚固的房间,问说最少需要多少栈灯。
解题思路:dfs+剪枝,暴力枚举放特殊灯的位置,然后将脆弱房间按照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,只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;
}
hdu 4770 Lights Against Dudely(回溯)
时间: 2024-10-11 22:49:59