Codevs 2800 tsp problem of sending and selling out

2800 take-out

time limit: 2 sspace limit: 256000 KBtitle level: Diamonds DiamondTitle 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.

Output description Output Description

A positive integer representing the minimum time spent

Sample input Sample Input

30 1 10 101 0 1 210 1 0 1010 2 10 0

Sample output Sample Output

8

Data range and Tips Data Size & Hint

1<=n<=15

Idea: Floyd Shortest path + state compression primer;

To die, starting from 1;

#include <bits/stdc++.h>using namespacestd;#definell Long Long#defineESP 0.00000000001Const intn=2e5+Ten, m=1e6+Ten, inf=1e9+Ten, mod=1e9+7;intdp[n][ -];intmp[ -][ -];intMain () {intx,y,z,i,t;  for(i=0; i<n;i++)         for(t=0;t< -; t++) Dp[i][t]=inf; scanf ("%d",&x); X++;  for(i=1; i<=x;i++)         for(t=1; t<=x;t++) scanf ("%d",&mp[i][t]);  for(intK =1; K <= x; k++)         for(inti =1; I <= x; i++)             for(intj =1; J <= X; J + +) Mp[i][j]= Min (Mp[i][j], mp[i][k] +Mp[k][j]); dp[2][1]=0;  for(t=0;t< (1<< (x+1)); t++)    {         for(i=1; i<=x;i++)        {             for(intj=1; j<=x;j++)            if((t& (1<<i))! =0) && (t& (1&LT;&LT;J))! =0)) Dp[t][i]=min (dp[t][i],dp[t^ (1<<i)][j]+Mp[j][i]); }    }    intans=inf;  for(intI=1; i<=x;i++) ans=min (ans,mp[i][1]+dp[(1<< (x+1))-2][i]); printf ("%d\n", ans); return 0;}

Codevs 2800 tsp problem of sending and selling out