题意：有1*2的小方块，在n*m （n <= 70, m <= 7）的棋盘内有一些点不能覆盖，要求将方块最少数量地放入棋盘，使得空格子两两不相邻。
m那么小，很明显是状态压缩DP对吧。
我们用f[i][j]表示在第i行，覆盖前一部分后当前状态为j的最小花费，转移是很显然的。重点是状态之间的关系如何确定，否则无法进行转移。考虑预处理出融洽的两个状态。用1表示在当前位放置了一块方块，我们从0枚举到(1 << m) - 1，我们要判断两个决策是否融洽，但是问题来了：棋盘上有不可覆盖点！！其实无所谓吧，我们可以尝试不预处理，而是在转移时枚举状态。复杂度：O(n*s*s)。好像没了？？2维数组真的够了么？？？当然不行啊！因为一个方块可以同时影响两行的状态，所以我们要改进状态表示，用三维数组吧。。。然后被细节问题坑成狗。空间卡的死，需要用滚动数组优化。dfs实现时，其实只要考虑好各种情况就行了。
1 //{HEADS 2 #define FILE_IN_OUT 3 #include <cstdio> 4 #include <cstring> 5 #include <cstdlib> 6 #include <cmath> 7 #include <ctime> 8 #include <algorithm> 9 #include <iostream> 10 #include <fstream> 11 #include <vector> 12 #include <stack> 13 #include <queue> 14 #include <deque> 15 #include <map> 16 #include <set> 17 #include <bitset> 18 #include <complex> 19 #include <string> 20 #define REP(i, j) for (int i = 1; i <= j; ++i) 21 #define REPI(i, j, k) for (int i = j; i <= k; ++i) 22 #define REPD(i, j) for (int i = j; 0 < i; --i) 23 #define STLR(i, con) for (int i = 0, sz = con.size(); i < sz; ++i) 24 #define STLRD(i, con) for (int i = con.size() - 1; 0 <= i; --i) 25 #define CLR(s) memset(s, 0, sizeof s) 26 #define SET(s, v) memset(s, v, sizeof s) 27 #define mp make_pair 28 #define pb push_back 29 #define PL(k, n) for (int i = 1; i <= n; ++i) { cout << k[i] << ' '; } cout << endl 30 #define PS(k) STLR(i, k) { cout << k[i] << ' '; } cout << endl 31 using namespace std; 32 void FILE_INIT(string FILE_NAME) { 33 #ifdef FILE_IN_OUT 34 #ifndef ONLINE_JUDGE 35 freopen((FILE_NAME + ".in").c_str(), "r", stdin); 36 freopen((FILE_NAME + ".out").c_str(), "w", stdout); 37 #endif 38 #endif 39 } 40 typedef long long LL; 41 typedef double DB; 42 typedef pair<int, int> i_pair; 43 const int INF = 0x3f3f3f3f; 44 const LL INFL = 0x3f3f3f3f3f3f3f3f; 45 //} 46 47 char gchar() { 48 char ret = getchar(); 49 for(; ret != '.' && ret != '*'; ret = getchar()); 50 return ret; 51 } 52 53 const int maxn = 74; 54 const int maxs = (1 << 7) + 3; 55 int G[maxn]; 56 int n, m, now, last; 57 58 LL f[2][maxs][maxs]; 59 int Snum; 60 61 //最好玩的部分开始！！ 62 void dfs(int col, int pre_s, int now_s, int next_s, int cnt, LL f_last) { 63 f(0 < col && !(pre_s & (1 << (col - 1))) && !(now_s & (1 << (col - 1)))) { 64 return ; 65 } 66 if(1 < col && !(now_s & (1 << (col - 1))) && !(now_s & (1 << (col - 2)))) { 67 return ; 68 } 69 if(col == m) { 70 f[now][now_s][next_s] = min(f[now][now_s][next_s], f_last + cnt); 71 return ; 72 } 73 dfs(col + 1, pre_s, now_s, next_s, cnt, f_last); 74 if(!(now_s & (1 << (col))) && !(next_s & (1 << (col)))) { 75 dfs(col + 1, pre_s, now_s | (1 << (col)), next_s | (1 << (col)), cnt + 1, f_last); 76 } 77 if(col + 1 < m && !(now_s & (1 << col)) && !(now_s & (1 << (col + 1)))) { 78 dfs(col + 1, pre_s, now_s | (1 << (col)) | (1 << (col + 1)), next_s, cnt + 1, f_last); 79 } 80 } 81 int main() { 82 FILE_INIT("132"); 83 84 scanf("%d%d", &n, &m) 85 for(int i = 1; i <= n; ++i) { 86 for(int j = 0; j < m; ++j) { 87 char c = gchar(); 88 if(c == '*') { 89 G[i] += 1 << j; 90 } 91 } 92 } 93 Snum = (1 << m) - 1; 94 now = 0, last = 1; 95 memset(f[now], INF, sizeof f[now]); 96 f[0][Snum][G[1]] = 0; 97 for(int i = 1; i <= n; ++i) { 98 swap(now, last); 99 memset(f[now], INF, sizeof f[now]); 100 for(int j = 0; j <= Snum; ++j) { 101 for(int k = 0; k <= Snum; ++k) { 102 if(f[last][j][k] < INFL) { 103 dfs(0, j, k, G[i + 1], 0, f[last][j][k]); 104 } 105 } 106 } 107 } 108 LL ans = INFL; 109 for(int i = 0; i <= Snum; ++i) { 110 ans = min(ans, f[now][i][0]); 111 } 112 printf("%lld ", ans); 113 return 0; 114 }
送测试数据：
3 2 .. .. ** 4 4 .**. *..* *..* .**. 7 7 ....... ....... ....... ...*... ....... ....... ....... 8 1 . . . . . . . . 1 1 . 3 3 *.* ... *.* 3 7 *.*.*.. ..*...* *...*.* 2 7 .....*. **..*.* 1 7 *.*.*.. 2 2 *. .* 70 7 ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... ....... 10 3 ..* ..* ..* ..* ..* *.. *.. *.. *.. *.. 10 4 *..* *..* *..* *..* *..* *..* *..* *..* *..* *..* 10 3 ..* ..* ..* ..* ..* ..* ..* ..* ..* ..* 10 2 .. .. .. .. .. .. .. .. .. .. 5 5 .*..* *.... ..**. **.*. .**..
2 2 16 3 0 1 3 2 1 0 164 7 7 7 7 4