UVA 1252 Twenty Questions 状压DP

简单状压DP:

当前状态s如果这个物品有状态a个属性,枚举下一个要猜测的特征k

dp[s][a]=min(dp[s][a],max(dp[s+k][a],dp[s+k][a+k])+1);

4643 - Twenty Questions

Asia - Tokyo - 2009/2010

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" orno”. 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" orno” 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" orno”) 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

4643 - Twenty Questions 1/301110100111

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

/* ***********************************************
Author        :CKboss
Created Time  :2015年02月22日 星期日 14时49分48秒
File Name     :UVA1252.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn=3000;

int n,m;
int num[300];
int dp[maxn][maxn];
int cnt[maxn][maxn];

void init()
{
    memset(dp,-1,sizeof(dp));
    memset(cnt,0,sizeof(cnt));
}

void dfs(int s,int a) /// 已经枚举特征s,w具有特征a
{
    if(dp[s][a]!=-1) return ;
    if(cnt[s][a]<=1)
    {
        dp[s][a]=0; return ;
    }
    for(int i=0;i<m;i++)
    {
        if(s&(1<<i)) continue ;
        int ns=s|(1<<i);
        int na=a|(1<<i);
        dfs(ns,a); dfs(ns,na);
        if(dp[s][a]==-1)
            dp[s][a]=max(dp[ns][a],dp[ns][na])+1;
        else
            dp[s][a]=min(dp[s][a],max(dp[ns][a],dp[ns][na])+1);
    }
}

int main() {
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    while(scanf("%d%d",&m,&n)!=EOF)
    {
        if(n==0&&m==0) break;
        init();
        for(int i=0;i<n;i++)
        {
            char temp[20];
            int ans=0;
            scanf("%s",temp);
            for(int j=0;j<m;j++)
                if(temp[j]==‘1‘) ans|=(1<<j);
            num[i]=ans;
        }
        for(int i=0;i<(1<<m);i++)
            for(int j=0;j<n;j++)
                cnt[i][num[j]&i]++;
        dfs(0,0);
        printf("%d\n",dp[0][0]);
    }
    return 0;
}