POJ 2288 Hamilton Loop DP Solution

Main topic:

There are n islands, so VI is the weighted value of the Island CI. A Hamilton path C1,C2,C3 ... The value of CN is 3 parts

In the first part, the weights of the islands in the path are added, and the second part (CI,CJ) on each side, plus all the VI*VJ

The third part, if 3 consecutive cities can form a 3-point unicom, then add VI*VJ*VK

Find a path that maximizes its weight and record how many paths can be reached for the maximum weight value

1->2->3, 3->2->1 as the same path

There are up to 13 cities, so use 2 to indicate whether to reach the current location of the city

Here Dp[i][k][j] means that the last two cities to reach the I state are j,k, so that the maximum weight that can be reached

Dp[i| ( 1<< (t-1)][j][t] = max{dp[i| ( 1<< (t-1)][j][t], dp[i][k][j]+val[t]+val[j]*val[t]+ edge[k][t]?val[j]*val[k]*val[t]:0}

Note that each time the DP value is updated, the number of paths that reach the current state is recorded cnt[i][k][j]

It's important to note that there is only one city to judge.

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <cmath>5 using namespacestd;6 #defineN 147 #definell Long Long8 intn,m;9ll dp[1<<n][n][n], Val[n], cnt[1<<N][N][N];//CNT records the total number of possible methods in the current stateTen BOOLEdge[n][n]; One  A voidGETDP () - { -memset (CNT,0,sizeof(CNT)); theMemset (DP,-1,sizeof(DP)); -     intAll = (1<<n); -dp[0][0][0]=0, cnt[0][0][0]=0; -      for(intI=1; I<=n; i++) dp[1<< (I-1)][0][i]=val[i],cnt[1<< (I-1)][0][i]=1; +      for(intI=1; I<all; i++){ -          for(intj=1; J<=n; J + +){ +             if(! (i& (1<< (J-1))))Continue; A              for(intk=0; K<=n; k++){ at                 if(k = =0&& (i!= (1<< (J-1))))Continue; -                 if(K==j | | (! (i& (1<< (K-1))) && k!=0))Continue; -                 if(dp[i][k][j]<0)Continue; -  -                  for(intt=1; T<=n; t++){ -                     if(!edge[j][t] | | i& (1<< (t1)))Continue; inll v = dp[i][k][j]+val[t]+val[t]*Val[j]; -                     if(Edge[t][k]) v = v+val[j]*val[k]*Val[t]; to                     intStatus = i| (1<< (t1)); +                     if(dp[status][j][t]<0|| dp[status][j][t]<v) { -dp[status][j][t]=v; thecnt[status][j][t]=Cnt[i][k][j]; *                     } $                     Else if(Dp[status][j][t] = =v) {Panax Notoginsengcnt[status][j][t]+=Cnt[i][k][j]; -                     } the                                         //Debug +                  //print (status);cout<<endl; A                    //cout<<i<< "<<j<<" "<<k<<" "<<dp[status][k][t]<<endl; the                 } +             } -         } $     } $ } -  - intMain () the { -    //freopen ("a.in", "R", stdin);Wuyi     intT; thescanf"%d", &T); -      while(t--) Wu     { -scanf"%d%d", &n, &m); About          for(intI=1; I<=n; i++) scanf ("%i64d", val+i); $Memset (Edge,0,sizeof(Edge)); -         intb; -          for(intI=0; I<m; i++){ -scanf"%d%d", &a, &b); Aedge[a][b]=true; +edge[b][a]=true; the         } - GETDP (); $         intAll= (1<<n); thell MAXN =-1, ans=0; the          for(intI=0; I<=n; i++) the              for(intj=0; J<=n; J + +) the             { -Maxn=max (maxn,dp[all-1][i][j]); in             } the         if(MAXN = =-1){ thePuts"0 0"); About             Continue; the         } the          for(intI=0; I<=n; i++) the              for(intj=0; J<=n; J + +) +                 if(MAXN = = dp[all-1][I][J]) ans+=cnt[all-1][i][j]; -         /** * To be aware that only one city is in the case of a special sentence **/ theprintf"%i64d%i64d\n", MAXN, ans/2? ans/2:1);Bayi     } the     return 0; the}

POJ 2288 Hamilton Loop DP Solution