State compression, when our state too much can be considered with bit to store, binary to represent the set, with & to take the intersection, with ^ to different or.
The DP process is very simple, traversing all cases to take the shortest path, because the shortest Hamiltonian loop itself is an NPC problem, the efficiency is not high.
#include <vector>#include<iostream>using namespacestd;//NP-Complete problem of the shortest Hamiltonian loop ... intmap[ -][ -]={0};intn=0;Const intinf=768000;//3000*16*16intf[1<< -][ -]={0};//F[i][j] represents the shortest path length from the start point through all the points in I to JvoidInitialize () {cin>>N; for(inti =0; I < n; ++i) { for(intj=0; J < N; ++j) {cin>>map[i][j];//input Data } }} intbuild () {//The starting point is always 1 in the list, so don't think about the starting point.//DP state compression such as n=4 eg:1<<3---1000 from 0000 to 1000 determines all cases in the previous 3 locations for(inti =0; I < (1<< (n1) ); ++i)// { //Note that the point labeled J in the input is the subscript for the j-1 abbreviation J point in I//DP calculation F[i][j] for(intj =1; J < N; ++j) { if(i = = (1<< (J-1) ))//If I had just a J-point at this point,F[I][J] = map[0][J];//direct from start to J Else{//Otherwise you need DP if(I & (1<< (J-1) ))//J points in I{F[i][j]= INF;//start looking for its minimum value for(intK =1; K < n; ++k)if(K!=j and (I & (1<< (K-1) ) ) ) //Find K K not J and in I//The state transition equation is actually just a relaxation operation. EdgeF[i][j] = min (f[i^ (1<< (J-1))][k] +Map[k][j], f[i][j]); } } } } intres =INF; for(inti =1; I < n; ++i) {res= Min (res, f[(1<< (n1))-1][i] + map[i][0]); } returnRes;}intMainintargcChar Const*argv[]) {Initialize (); if(n==1) cout<<2*map[0][0]<<Endl; Elsecout<<build () <<Endl; return 0;}
"Algorithmic Learning Notes" 62. State compression DP SJTU OJ 1088 postman small F