[TSP+floyd]POJ3311 Hie with the Pie

题意: 给i到j花费的地图 1到n编号   一个人要从1遍历n个城市后回到1

求最小的花费(可以重复走)

分析

http://www.cnblogs.com/Empress/p/4039240.html

TSP

因为可以重复走 所以先floyd一下求最短路

 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cmath>
 #include <string>
 #include <sstream>
 #include <iostream>
 #include <algorithm>
 #include <iomanip>
 using namespace std;
 #include <queue>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <set>
 #include <map>
 typedef long long LL;
 typedef long double LD;
 #define pi acos(-1.0)
 #define lson l, m, rt<<1
 #define rson m+1, r, rt<<1|1
 typedef pair<int, int> PI;
 typedef pair<int, PI> PP;
 #ifdef _WIN32
 #define LLD "%I64d"
 #else
 #define LLD "%lld"
 #endif
 //#pragma comment(linker, "/STACK:1024000000,1024000000")
 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
 //inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}
 //inline void print(LL x){printf(LLD, x);puts("");}
 //inline void read(double &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}
 
 int dp[<<][], mp[][];
 int n;
 void floyd()
 {
     for(int k=;k<n;k++)
         for(int i=;i<n;i++)
             for(int j=;j<n;j++)
                 mp[i][j]=min(mp[i][j], mp[i][k]+mp[k][j]);
 }
 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w", stdout);
 #endif
     while(~scanf("%d", &n) && n)
     {
         n++;
         for(int i=;i<n;i++)
             for(int j=;j<n;j++)
                 scanf("%d", &mp[i][j]);
         floyd();
         memset(dp, , sizeof(dp));
         dp[(<<n)-][]=;
         for(int s=(<<n)-;s>=;s--)
             for(int v=;v<n;v++)
                 for(int u=;u<n;u++)
                     if(!(s>> u & ))
                         dp[s][v]=min(dp[s][v], dp[s | <<u][u]+mp[v][u]);
         printf("%d\n", dp[][]);
     }
     return ;
 }

POJ 3311