Hie with the Pie (POJ 3311 pressure DP)

Test instructions: Pizza shop gives pizza to n places, the time it takes to spend between places (including the pizzeria), and the minimum time it takes to send all the places and back to the store

Analysis: The state is ok dp[i][j],i in 1 means that the place has been sent, otherwise 0,j for the current state of the last place to send, pay attention to how to walk only calculate the minimum time, of course, the shortest way, point rarely by Floyd to find out the most short-circuited points

For the minimum time to return to the store, depart from the store (status 1).

#include <map> #include <set> #include <list> #include <cmath> #include <queue> #include <stack> #include <cstdio> #include <vector> #include <string> #include <cctype> #include <complex> #include <cassert> #include <utility> #include <cstring> #include <cstdlib># Include <iostream> #include <algorithm>using namespace std;typedef pair<int,int> pii;typedef long Long ll; #define Lson l,m,rt<<1#define Pi ACOs ( -1.0) #define Rson m+1,r,rt<<11#define all 1,n,1#define read fre Open ("In.txt", "R", stdin) const LL INFLL = 0x3f3f3f3f3f3f3f3fll;const int inf= 0x7ffffff;const int mod = 1000000007;int    D[15][15],a[15][15],n,dp[2500][15];void solve () {int c= (1<< (n+1))-1; for (int k=0;k<=n;++k) for (int. i=0;i<=n;++i) for (int j=0;j<=n;++j) d[i][j]=min (d[i][j],d[i][k    ]+D[K][J]);    Memset (Dp,-1,sizeof (DP));    dp[1][0]=0;  for (int i=1;i<=c;++i) {i|=1;  for (int j=0;j<=n;++j) {if (dp[i][j]==-1) continue; for (int k=0;k<=n;++k) if (dp[i| ( 1<<k)][k]==-1&&j!=k| | Dp[i| ( 1<<k)][k]>dp[i][j]+d[j][k]) dp[i| (    1<<k)][k]=dp[i][j]+d[j][k]; }} printf ("%d\n", Dp[c][0]);}        int main () {while (~SCANF ("%d", &n)) {if (n==0) break;        for (int i=0;i<=n;++i) for (int j=0;j<=n;++j) scanf ("%d", &d[i][j]);    Solve (); }return 0;}

　　

Hie with the Pie (POJ 3311 pressure DP)