codevs2800 Delivery of Takeaway

• Title Description Description
  There was a delivery, he had n orders on his hand, and he had to deliver n parts to the hands of n different clients. N different customers are in 1~n numbered cities. Delivery from the city of No. 0, and then N cities to walk once (a city can go many times), and finally back to 0 points (his unit), ask the shortest time is how much. The time for direct access to any two cities is now known.

• Enter a description input Description
  First line a positive integer n (1<=n<=15)
  Next is a (n+1) * (n+1) matrix in which the number in the matrix is a positive integer that does not exceed 10000. The I row j column of the Matrix represents the time of the direct access between City I-1 and City j-1. Of course, the direct access time of city A to City B and City B to city A is not necessarily the same, that is, the road is one-way.

• outputs description output Description
  A positive integer representing the minimum time spent

• sample input to sample
  3
  0 1 10 10
  1 0 1 2
  10 1 0 10
  10 2 10 0

• sample output Sample outputs
  8

• Data Size & Hint
  1<=n<=15

• Solving
  A see can go many times first want to use Floyd to find the shortest way, N only 15~
  Below is the pressure DP. (For the first time, it's not easy to think about all the states and to design a correct transfer equation.)
  We make F[i][lst] show the first state when I went to the first LST city to spend the minimum cost, wherein I converted to binary m, with an array of p[] records M 1 position, then I state is p[]-1 all cities have traversed and other cities have not traversed the state, so lst+ The position of 1 in M is 1 o'clock, the boundary condition F[0][0]=0, the remainder is the INF, and the final state is f[(1 << (n+1) -1][0].

• Code

#include <cstdio>#include <algorithm>#include <cstring>using namespace STD;Const intMAXN = -, oo =1000000000;int Map[MAXN] [MAXN], f[1<< MAXN][MAXN], N;inline voidInit () {scanf("%d", &n); for(inti =0; I <= N; ++i) { for(intj =0; J <= N; ++J) {scanf("%d", &MapI        [j]); }    }//Below is Floyd     for(intK =0; K <= N; ++K) { for(inti =0; I <= N; ++i) { for(intj =0; J <= N; ++J) {MapI [j] = min (MapI [j],MapI [K] +MapK            [j]); }        }    }}inline voidWork () {memset(F,127/3,sizeof(f)); f[0][0] =0; for(inti =0; I < (1<< (n +1)); ++i) { for(intLST =0; LST <= N; ++LST) {if((I | (1<< lst)) = = i)//Last to the city in the city that has been visited, the state is legal{ for(intPre =0; Pre <= N; ++pre) {F[i][lst] = min (F[i][lst], F[i-(1<< lst)][pre] +Map[Pre] [LST]);//has not been to LST before, now through the pre to LSTF[i][lst] = min (F[i][lst], F[i][pre] +Map[Pre] [LST]);//Before the LST, now through the pre and to the LST}            }        }    }printf("%d", f[(1<< (n +1)) -1][0]);}intMain () {init (); Work ();return 0;}

codevs2800 Delivery of Takeaway