UVA 1252 Twenty Questions pressure DP

Simple pressure DP:
Current state S If this item has a status of a attribute, enumerate the next feature K to guess
d P[s][a]=mIN(d P[s][a],max(d P[s+k][a],d P[s+k][a+k])+1);

4643-twenty Questions

asia-tokyo-2009/2010
Consider a closed world and a set of features that is defined for all the objects in the world. Each feature can
Be answered withyes" orNo ". Using those features, we can identify any object from the rest of the objects
In the world. In the other words, each object can be represented as a fixed-length sequence of booleans. Any
object is different from and objects by at least one feature.
You would a like to identify a object from others. For the purpose, you can ask a series of questions to
Someone who knows what's the object is. Every question you can ask was about one of the features. He/she
Immediately answers each question withyes" orNo "correctly. You can choose the next question
Get the answer to the previous question.
You kindly pay the answerer of yen as a tip for each question. Because you don ' t has surplus money, it's
necessary to minimize the number of questions in the worst case. You don ' t know, the correct answer,
But fortunately know all the objects in the world. Therefore, you can plan a optimal strategy before you start
Questioning.
The problem you has 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 integers, m
and n:the number of features, and the number of objects, respectively. You can assume 0 < m one and 0 <
N 128. It is followed by n lines and each of the which corresponds to an object. Each line includes a binary string of
Length m which represent the value (yes" orNo ") of features. There is no identical objects.
The end of the input is indicated by a line containing the zeros. There is at the most 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:ckbosscreated time:2015 February 22 Sunday 14:49 48 seconds file Na me: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 intmaxn= the;intN,m;intnum[ -];intDP[MAXN][MAXN];intCNT[MAXN][MAXN];voidInit () {memset(dp,-1,sizeof(DP));memset(CNT,0,sizeof(CNT));}voidDfsintSintA/// already enumerated feature s,w with feature a{if(dp[s][a]!=-1)return;if(cnt[s][a]<=1) {dp[s][a]=0;return; } for(intI=0; i<m;i++) {if(s& (1<<i))Continue;intns=s| (1<<i);intna=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;ElseDp[s][a]=min (Dp[s][a],max (Dp[ns][a],dp[ns][na]) +1); }}intMain () {//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(intI=0; i<n;i++) {Chartemp[ -];intans=0;scanf('%s ', temp); for(intj=0; j<m;j++)if(temp[j]==' 1 ') ans|= (1&LT;&LT;J);        Num[i]=ans; } for(intI=0;i< (1&LT;&LT;M); i++) for(intj=0; j<n;j++) cnt[i][num[j]&i]++; Dfs0,0);printf("%d\n", dp[0][0]); }return 0;}

UVA 1252 Twenty Questions pressure DP