Algorithm: UVa 1252

The main effect of the topic

There are n binary strings of length m, each of which is different.

In order to separate all the strings, you can ask whether a bit is 0 or 1 at a time.

Ask the least number of questions, and you can distinguish all strings.

Ideas

F[S1][S2]: The question is {S1} collection, the answer is {s2}, you need to ask several times before you can separate all areas

When the problem set is {S1}, if you can't distinguish all the answers, you need to continue to ask another question.

Then you can launch the next set of questions:

Nextquestions = {S1 | (1<<k) when the K-bit of the S1 is 0}

Then you can get:

F[S1][S2] = 0, if the same number as the answer s2 is less than or equal to 1, then all can be separated, and ask 0 times

F[S1][S2] = {min (f[nextquestions][s1], f[nextquestions][s1^ (1<<k)), when S1 's K-bit is 0 o'clock}

Code

/**========================================== * is a solution for ACM/ICPC problem * * @source: uva-1252 Twenty Q Uestion * @type: Memory search * @author: Shuangde * @blog: blog.csdn.net/shuangde800 * @email: zengshuangde@gmail.com *========== =================================*/#include <iostream> #include <cstdio> #include <algorithm> #
    
include<vector> #include <queue> #include <cmath> #include <cstring> using namespace std;
typedef long long Int64;
const int INF = 0X3F3F3F3F;
    
Const double PI = ACOs (-1.0);
const int MAXN = 130;
    
int n, m;
int P[MAXN];
int f[(1<<11) +10][(1<<11) +10];
    
www.bianceng.cn int dfs (int s1, int s2) {if (F[S1][S2)!= INF) return F[S1][S2];
int cnt = 0;
    
for (int i = 0; i < n; ++i) if ((P[i] & S1) = = s2) ++cnt;
    
if (CNT <= 1) {f[s1][s2] = 0; return 0;} for (int i=0; i<m; ++i) {if (s1& (1<<i)) continue; int nextquest = S1 |
(1<<i); int& d = f[s1][s2];
d = min (d, Max (Dfs (nextquest, S2), DFS (Nextquest, s2^ (1<<i))) +1);
return F[S1][S2];  int main () {char str[130]; while (~SCANF ("%d%d", &m, &n) && n+m) {for (int i = 0; i < n; ++i) {scanf ("%s", str); p[i] = 0; for (int j = 0; Str[j] ++j) if (str[j] = = ' 1 ') p[i] |= 1<<j;} memset (f, INF, size
Of (f));
printf ("%d\n", DFS (0, 0));
return 0; }