UVA 1252-twenty Questions (state compression dp+ subset enumeration)

Main topic: There are n items, each item has m characteristics, each item may have or not each feature, now assume that an item, by asking certain characteristics to determine the item, ask the maximum number of times you can determine the item.

Different strategies for further questioning may be taken after each inquiry, depending on the answer.

With D[S][S0], the answer is S0 (that is, the item has S0 in s), at least the number of times it needs to be asked. Enumerates the feature completion recursion for the next query. The final d[0][0] is the answer. S0 is obviously a subset of S. The next question is not characteristic of S. If there is only one item for a D[S][S0], then the value is 0 and no more queries are required.

State transition equation:

d[s][s0]=min {max (d[s|u][s0],d[s|u][s0|u])} (u== (1<<k), s&u==0)

#include <stdio.h> #include <stdlib.h>int a[150];int d[2100][2100];char e[1010];int Main (void) {int i,j,u,p , q,n,m,sump,minp,ok,s,s0;scanf ("%d%d", &m,&n); while ((m!=0) | | (n!=0)) {for (i=1;i<=n;i++) {scanf ("%s", e+1), Sump=0;for (j=1;j<=m;j++) {sump=2*sump+e[j]-' 0 ';} A[i]=sump;} p= (1<<m) -1;for (s=p; s>=0; s--) {S0=s;while (1) {sump=0;ok=1;for (i=1;i<=n;i++) {if ((a[i]&s) ==s0) {sump++;if (sump>1) {OK=0;break;}}} if (ok==1) {d[s][s0]=0;} Else{minp= (1<<10); for (i=1;i<=m;i++) {u=1<< (i-1); if ((U&s) ==0) {q=d[s|u][s0]>d[s|u][s0|u]?d[ S|U][S0]:d [S|u][s0|u];minp=q<minp?q:minp;}} d[s][s0]=minp+1;} if (s0==0) {break;} s0= (s0-1) &s;}} printf ("%d\n", D[0][0]); scanf ("%d%d", &m,&n);} return 0;}

UVA 1252-twenty Questions (state compression dp+ subset enumeration)