UVA 1252 (State compression DP)

Test instructions: There are n binary strings, the length is m and all are not the same, ask at least how many questions can be divided into a complete number of n string.
The problem: 1<=m<=11, from this range can be inferred is the state compression, then DP must have one dimension to ask questions, and then another is based on the proposed problem of the string classification, one is in line with the state of the proposed problem, the other is not in accordance with. This f[i][j] indicates how many questions to ask when the answer is state j in question I, in order to separate all the strings.
If found in question I and Answer j the same string only 1 string or not, the description f[i][j]=0 do not need to ask questions to have been differentiated, otherwise you will ask questions, the previous question I is 0 to 1, the answer in the bit 1 and remain unchanged DP return value to choose a larger value ( Because the two sides to distinguish between the open, choose the number of questions can be more.

#include <stdio.h>#include <string.h>#include <vector>#include <algorithm>using namespace STD;Const intINF =0x3f3f3f3f;Const intN =135;Const intM = (1<< One) +5;intN, M, A[n], f[m][m];Charstr[ the];intdpintS1,intS2) {if(F[S1][S2]! = INF)returnF[S1][S2];intCNT =0; for(inti =0; I < n; i++)if((S1 & a[i]) = = s2) cnt++;if(CNT <=1)returnF[S1][S2] =0; for(inti =0; I < m; i++) {if(S1 & (1<< i))Continue;inttemp = S1 | (1<< i); F[S1][S2] = min (F[s1][s2], Max (DP (temp, S2), DP (temp, s2 ^ (1<< i)) +1); }returnF[S1][S2];}intMain () { while(scanf("%d%d", &m, &n) = =2&& n + m) {memset(F, INF,sizeof(f)); for(inti =0; I < n; i++) {scanf('%s ', str); A[i] =0; for(intj =0; J < M; J + +)if(Str[j] = =' 1 ') A[i] |= (1<< j); }printf("%d\n", DP (0,0)); }return 0;}

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UVA 1252 (State compression DP)