UVA1663 Purifying Machine

The decrease in the number of template collections is due to matching pairs of strings that are not matched in the string collection. So just find the maximum match for a binary graph.

The code does not explicitly create a binary diagram, you can imagine two sets X and Y, because two sets will run again, so a match will be counted two times, the return time divided by 2 on the line.

Pick up the Hungarian algorithm that almost forgot ...

#include <bits/stdc++.h>using namespacestd;Const intMAXN =Ten;Const intN =1<<MAXN;#defineDebug (x) \Bitset< ->BS (x); cout<<bs<<Endl;BOOLVis[n];intMatch[n];intDfst[n], Dfstime, N, M, M,tot;BOOLDfsintu) {     for(inti =0; I < n; i++){        intv = u^ (1<<i); if(Vis[v] && dfst[v]! =dfstime) {Dfst[v]=Dfstime; if(!~match[v] | |DFS (Match[v])) {Match[v]=u; return true; }        }    }    return false;}intMaxmatch () {memset (match,-1,sizeof(int)*M); memset (DFST,0,sizeof(int)*M); Dfstime=0; intAns =0;  for(inti =0; i < M; i++){        if(Vis[i]) {dfstime++; if(Dfs (i)) ans++; }    }    returnAns>>1;}intRead () {M=1<<N; memset (Vis,0,sizeof(BOOL)*M);  for(inti =0; I < m; i++){        intFG =-1, t =0; scanf ("\ n"); Charch;  for(intj =0; J < N; J + +) {ch=GetChar (); if(ch = ='*') FG =J; if(ch! ='0') T |=1<<J; } Vis[t]=true; if(~FG) vis[t^ (1&LT;&LT;FG)] =true; }    intRET =0;  for(inti =0; i < M; i++)if(Vis[i]) ret++; returnret;}intMain () {//freopen ("In.txt", "R", stdin);     while(SCANF ("%d%d",&n,&m), N) {printf ("%d\n", read ()-Maxmatch ()); }    return 0;}

UVA1663 Purifying Machine