[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

 

 1 #include <cstdio>

 2 #include <cstdlib>

 3 #include <cstring>

 4 #include <climits>

 5 #include <cctype>

 6 #include <cmath>

 7 #include <string>

 8 #include <sstream>

 9 #include <iostream>

10 #include <algorithm>

11 #include <iomanip>

12 using namespace std;

13 #include <queue>

14 #include <stack>

15 #include <vector>

16 #include <deque>

17 #include <set>

18 #include <map>

19 typedef long long LL;

20 typedef long double LD;

21 #define pi acos(-1.0)

22 #define lson l, m, rt<<1

23 #define rson m+1, r, rt<<1|1

24 typedef pair<int, int> PI;

25 typedef pair<int, PI> PP;

26 #ifdef _WIN32

27 #define LLD "%I64d"

28 #else

29 #define LLD "%lld"

30 #endif

31 //#pragma comment(linker, "/STACK:1024000000,1024000000")

32 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}

33 //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;}

34 //inline void print(LL x){printf(LLD, x);puts("");}

35 //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();}}

36 

37 int dp[1<<11][11], mp[11][11];

38 int n;

39 void floyd()

40 {

41     for(int k=0;k<n;k++)

42         for(int i=0;i<n;i++)

43             for(int j=0;j<n;j++)

44                 mp[i][j]=min(mp[i][j], mp[i][k]+mp[k][j]);

45 }

46 int main()

47 {

48 #ifndef ONLINE_JUDGE

49     freopen("in.txt", "r", stdin);

50     freopen("out.txt", "w", stdout);

51 #endif

52     while(~scanf("%d", &n) && n)

53     {

54         n++;

55         for(int i=0;i<n;i++)

56             for(int j=0;j<n;j++)

57                 scanf("%d", &mp[i][j]);

58         floyd();

59         memset(dp, 127, sizeof(dp));

60         dp[(1<<n)-1][0]=0;

61         for(int s=(1<<n)-2;s>=0;s--)

62             for(int v=0;v<n;v++)

63                 for(int u=0;u<n;u++)

64                     if(!(s>> u & 1))

65                         dp[s][v]=min(dp[s][v], dp[s | 1<<u][u]+mp[v][u]);

66         printf("%d\n", dp[0][0]);

67     }

68     return 0;

69 }

POJ 3311