UVa 1252 Twenty Questions

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 fromthe 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 thecorrect 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 numberof features, and the number of objects, respectively. You can assume 0 < m11 and 0< n128. It is followed by n lines, each of which corresponds to an object. Eachline 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

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <cmath>using namespace std;// 问题数目，对象数目int num_q, num_obj;// 每个对象的特征// array[i]的第j位为1代表第i个对象具有j特征int array[150];// cnt[s][a]代表询问s特征集合，满足a特征集合(不满足s-a特征集合)的对象个数int cnt[2200][2200];// record[s][a]代表已询问s特征集合，目标对象满足a特征集合(不满足s-a特征集合)的最少问题个数int record[2200][2200];int get_min(int s, int a);int main(){while((cin >> num_q >> num_obj) && !(num_q == 0 && num_obj == 0)){memset(record, -1, sizeof(record));memset(cnt, 0, sizeof(cnt));// 读入情况for(int i = 1; i <= num_obj; i++){string s;cin >> s;int num = 0;for(int j = 0; j < s.length(); j++)num = num*2 + (s[j]-'0');array[i] = num;//			cout << "s: " << s << " array: " << array[i] << endl;}// 计算cntint end = pow(2, num_q)-1;for(int s = 0; s <= end; s++){for(int i = 1; i <= num_obj; i++){int a = s & array[i];cnt[s][a]++;}}// 计算结果printf("%d\n", get_min(0, 0));}return 0;}// 计算已询问s特征集合，目标对象满足a特征集合(不满足s-a特征集合)的最少问题个数结果int get_min(int s, int a){if(record[s][a] != -1)return record[s][a];if(cnt[s][a] == 1){record[s][a] = 0;return record[s][a];}// 检查所有可行的继续问的特征record[s][a] = (1<<30);for(int i = 1; i <= num_q; i++){int this_s = (1 << (num_q-i));if((this_s & s) == 0){int r1 = get_min(s+this_s, a) + 1;int r2 = get_min(s+this_s, a+this_s) + 1;int r = max(r1, r2);if(r < record[s][a])record[s][a] = r;}}return record[s][a];}

这题一开始想是构造状态d(c, s)代表选了问题集合c以后，现在需要分的对象集合s，该情况下还需要问的最少问题。

后来发现s可以达到2^128，无法实现。

看了书，发现的做法是假设一个对象，d(s, a)代表选了问题集合s, 该对象符合问题a集合，该情况下还需要问的最少问题。

另外，这个问题就是构造最优的决策树，动态规划得到的答案复杂度是特征指数级的。构造最优决策树问题上世纪70年代就被证明是NPC了。

所以现在的算法都是基于启发式来贪心(ID3,C4.5等等)。