状压DP+记忆化搜索 UVA 1252 Twenty Questions

题目传送门

 /*
     题意：给出一系列的01字符串，问最少要问几个问题(列)能把它们区分出来
     状态DP+记忆化搜索：dp[s1][s2]表示问题集合为s1。答案对错集合为s2时，还要问几次才能区分出来
             若和答案(自己拟定)相差小于等于1时，证说明已经能区分了，回溯。否则还要加问题再询问
 */
 /************************************************
 * Author        :Running_Time
 * Created Time  :2015-8-13 10:54:26
 * File Name     :UVA_1252.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 MAXN =  + ;
 const int INF = 0x3f3f3f3f;
 const int MOD = 1e9 + ;
 int dp[(<<)+][(<<)+];
 int p[MAXN];
 char str[];
 int m, n;

 int DFS(int s1, int s2) {
     if (dp[s1][s2] != INF)  return dp[s1][s2];
     int cnt = ;
     for (int i=; i<=n; ++i)    {
         if ((p[i] & s1) == s2)  cnt++;
     }
     if (cnt <= )   {
         return dp[s1][s2] = ;
     }
     for (int i=; i<m; ++i) {
         if (s1 & ( << i))  continue;
         int nex = s1 | ( << i);
         dp[s1][s2] = min (dp[s1][s2], max (DFS (nex, s2), DFS (nex, s2 ^ ( << i))) + );
     }

     return dp[s1][s2];
 }

 int main(void)  {       //UVA 1252 Twenty Questions
     while (scanf ("%d%d", &m, &n) == ) {
         if (!m && !n)   break;
         for (int i=; i<=n; ++i)    {
             scanf ("%s", str);  p[i] = ;
             for (int j=; j<m; ++j)  {
                 if (str[j] == '')  {
                     p[i] |= ( << j);
                 }
             }
         }
         memset (dp, INF, sizeof (dp));
         printf ("%d\n", DFS (, ));
     }

     return ;
 }