[TSP+floyd]POJ3311 Hie with the Pie

时间：2023-03-08 20:26:22

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

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