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題目傳送門

題意：有n*m的房間，'.'表示可以被點亮，'#'表示不能被點亮，每點亮一個房間會使旁邊的房間也點亮，有意盞特別的燈可以選擇周圍不同方向的房間點亮。問最少需要多少燈使得所有房間點亮

分析：需要被點亮的房間最多只有15個，所以考慮狀壓，然后暴力枚舉選擇哪一個當(dāng)作特殊燈和枚舉選擇哪個方向使旁邊的房間亮，注意清空vis數(shù)組需要優(yōu)化，memset超時。上交6分鐘1Y，Orz。。。額，看錯榜了，最快的19分鐘，而且這不是第一道水題，汗

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/************************************************* Author :Running_Time* Created Time :2015/10/22 星期四 18:25:25* File Name :A.cpp************************************************/#include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std;#define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 2e2 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; const double EPS = 1e-8; char room[N][N]; struct Point {int x, y;Point (int x, int y) : x (x), y (y) {} }; int n, m; bool vis[N][N];void back_nomal(int x, int y) {vis[x][y] = false;if (x - 1 >= 1) vis[x-1][y] = false;if (y + 1 <= m) vis[x][y+1] = false; }void back_special(int x, int y, int type) {if (type == 1) {back_nomal (x, y); return ;}else if (type == 2) {if (x + 1 <= n) {vis[x+1][y] = false;}if (y + 1 <= m) {vis[x][y+1] = false;}}else if (type == 3) {if (x + 1 <= n) {vis[x+1][y] = false;}if (y - 1 >= 1) {vis[x][y-1] = false;}}else if (type == 4) {if (x - 1 >= 1) {vis[x-1][y] = false;}if (y - 1 >= 1) {vis[x][y-1] = false;}}vis[x][y] = false; }bool light_nomal(int x, int y) {if (x - 1 >= 1 && room[x-1][y] != '.') return false;if (y + 1 <= m && room[x][y+1] != '.') return false;if (x - 1 >= 1) {vis[x-1][y] = true;}if (y + 1 <= m) {vis[x][y+1] = true;}vis[x][y] = true;return true; }bool light_special(int x, int y, int type) {if (type == 1) {return light_nomal (x, y);}else if (type == 2) {if (x + 1 <= n && room[x+1][y] != '.') return false;if (y + 1 <= m && room[x][y+1] != '.') return false;if (x + 1 <= n) {vis[x+1][y] = true;}if (y + 1 <= m) {vis[x][y+1] = true;}}else if (type == 3) {if (x + 1 <= n && room[x+1][y] != '.') return false;if (y - 1 >= 1 && room[x][y-1] != '.') return false;if (x + 1 <= n) {vis[x+1][y] = true;}if (y - 1 >= 1) {vis[x][y-1] = true;}}else if (type == 4) {if (x - 1 >= 1 && room[x-1][y] != '.') return false;if (y - 1 >= 1 && room[x][y-1] != '.') return false;if (x - 1 >= 1) {vis[x-1][y] = true;}if (y - 1 >= 1) {vis[x][y-1] = true;}}vis[x][y] = true;return true; }int main(void) {while (scanf ("%d%d", &n, &m) == 2) {if (!n && !m) break;for (int i=1; i<=n; ++i) {scanf ("%s", room[i] + 1);}int cnt = 0;memset (vis, false, sizeof (vis));vector<Point> V;for (int i=1; i<=n; ++i) {for (int j=1; j<=m; ++j) {if (room[i][j] == '.') {cnt++;V.push_back (Point (i, j));}}}if (cnt == 0) {puts ("0"); continue;}int tot = 1 << cnt;int ans = INF;for (int i=0; i<tot; ++i) {for (int k=0; k<cnt; ++k) {for (int l=1; l<=4; ++l) {bool ok = true;for (int j=0; j<cnt; ++j) {if (i & (1 << j)) {if (j == k) {if (!light_special (V[j].x, V[j].y, l)) {ok = false; break;}}else {if (!light_nomal (V[j].x, V[j].y)) {ok = false; break;}}}}if (!ok) {for (int j=0; j<cnt; ++j) {if (j == k) back_special (V[j].x, V[j].y, l);else back_nomal (V[j].x, V[j].y);}continue;}bool flag = true;for (int j=0; j<cnt; ++j) {int x = V[j].x, y = V[j].y;if (!vis[x][y]) {flag = false; break;}}if (flag) {ans = min (ans, __builtin_popcount (i));}for (int j=0; j<cnt; ++j) {if (j == k) back_special (V[j].x, V[j].y, l);else back_nomal (V[j].x, V[j].y);}}}}printf ("%d\n", ans == INF ? -1 : ans);}return 0; }

　　

轉(zhuǎn)載于:https://www.cnblogs.com/Running-Time/p/4902677.html

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