UVa 1252 - Twenty Questions（记忆化搜索，状态压缩dp）

题目：

Consider a closed world and a set of features that are defined for all the objects in the world. Each
feature can be answered with “yes” or “no”. Using those features, we can identify any object from
the rest of the objects in the world. In other words, each object can be represented as a fixed-length
sequence of booleans. Any object is different from other objects by at least one feature.
You would like to identify an object from others. For this purpose, you can ask a series of questions
to someone who knows what the object is. Every question you can ask is about one of the features.
He/she immediately answers each question with “yes” or “no” correctly. You can choose the next
question after you get the answer to the previous question.
You kindly pay the answerer 100 yen as a tip for each question. Because you don’t have surplus
money, it is necessary to minimize the number of questions in the worst case. You don’t know what
is the correct answer, but fortunately know all the objects in the world. Therefore, you can plan an
optimal strategy before you start questioning.
The problem you have to solve is: given a set of boolean-encoded objects, minimize the maximum
number of questions by which every object in the set is identifiable.
Input
The input is a sequence of multiple datasets. Each dataset begins with a line which consists of two
integers, m and n: the number of features, and the number of objects, respectively. You can assume
0 < m ≤ 11 and 0 < n ≤ 128. It is followed by n lines, each of which corresponds to an object. Each
line includes a binary string of length m which represent the value (“yes” or “no”) of features. There
are no two identical objects.
The end of the input is indicated by a line containing two zeros. There are at most 100 datasets.
Output
For each dataset, minimize the maximum number of questions by which every object is identifiable and
output the result.
Sample Input
8 1
11010101
11 4
00111001100
01001101011
01010000011
01100110001
11 16
01000101111
01011000000
01011111001
01101101001
01110010111
01110100111
10000001010
10010001000
10010110100
10100010100
10101010110
10110100010
11001010011
11011001001
11111000111
11111011101
11 12
10000000000
01000000000
00100000000
00010000000
00001000000
00000100000
00000010000
00000001000
00000000100
00000000010
00000000001
00000000000
9 32
001000000
000100000
000010000
000001000
000000100
000000010
000000001
000000000
011000000
010100000
010010000
010001000
010000100
010000010
010000001
010000000
101000000
100100000
100010000
100001000
100000100
100000010
100000001
100000000
111000000
110100000
110010000
110001000
110000100
110000010
110000001
110000000
0 0
Sample Output
0
2
4
11
9

本文出自    http://blog.youkuaiyun.com/shuangde800

题目大意

有n个长度为m的二进制串，每个都是不同的。

为了把所有字符串区分开，你可以询问，每次可以问某位上是0还是1。

问最少提问次数，可以把所有字符串区分开来。

状态dp；

dp[s1][s2]表示 提问的问题是s1集合，答案是s2时，还需要问几次才可以全部区分开;

s1,s2都可以用一个m位的2进制状态来表示；

s1就是如果你要询问第几位，第几位就表示为一；

s2就是s1&a[i]的结果，可以表示所有的答案状态；

当问题集合为{s1}时， 如果还不能区分所有答案，那么就需要继续再问一个问题，
那么可以推出下一个问题的集合为：
nextQuestions = { s1 | (1<<k),  当s1的k位上为0的时候 }

那么可以得到：
f[s1][s2] = 0, 如果和答案s2相同的个数小于等于1，那么已经可以全部区分开了，还要询问0次
f[s1][s2] = { min(f[nextQuestions][s1], f[nextQuestions][s1^(1<<k)]), 当s1的k位上为0时}

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 1<<11;
int dp[MAXN][MAXN];
int a[133], n, m;
//备忘录搜索
int dfs(int s1, int s2) {
    if (dp[s1][s2] != INF) return dp[s1][s2];
    int cnt = 0;
    for(int i = 0; i < n; i++)
        if ((s1&a[i]) == s2) cnt++;
    if (cnt <= 1) return dp[s1][s2] = 0;
    if (cnt == 2) return dp[s1][s2] = 1;

    for(int i = 0; i < m; i++) {
        int tmp = 1<<i;
        if (s1&tmp) continue;
        dp[s1][s2] = min(dp[s1][s2], max(dfs(s1|tmp, s2), dfs(s1|tmp, s2|tmp))+1);
    }
    return dp[s1][s2];
}

int main() {
    char c[12];
    while (scanf("%d%d", &m, &n) && n+m) {
        for(int i = 0; i < n; i++) {
            scanf("%s", c);
            a[i] = 0;
            for(int j = 0; j < m; j++)
                if (c[j]-'0')
                    a[i] |= (1<<j);
        }
        memset(dp, INF, sizeof(dp));
        printf("%d\n", dfs(0, 0));
    }
    return 0;
}