状压DP Poj3311 Hie with the Pie

It is limited, at the problem is not the solution, a lot of understanding

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Problem C: Poj3311 Hie with the Pie

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 58 Solved: 29
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Description

The Pizazz Pizzeria prides itself in delivering pizzas to its customers as fast as possible. Unfortunately, due to cutbacks, they can afford to hire only one driver to do the deliveries. He will wait for 1 or more (up to 10) orders to be processed before he starts any deliveries. Needless to say, he would like to take the shortest route in delivering these goodies and returning to the pizzeria, even if it means passing the same location (s) or the pizzeria more than once on the way. He has commissioned you to write a program to help him.
you have a n + 1 (1 <= n <= 10) have a complete graph to the point, it is given in matrix form between any two different points distance. (Where the distance from i to j is not necessarily equal to the distance from j to i) want you to find leaves from 0:00, walked to the number n 1 point of at least once, and then return to spend the minimum number 0:00 distance

Input

Input will consist of multiple test cases. The first line will contain a single integer n indicating the number of orders to deliver, where 1 ≤ n ≤ 10. After this will be n + 1 lines each containing n + 1 integers indicating the times to travel between the pizzeria (numbered 0) and the n locations (numbers 1 to n). The jth value on the ith line indicates the time to go directly from location i to location j without visiting any other locations along the way. Note that there may be quicker ways to go from i to j via other locations, due to different speed limits, traffic lights, etc. Also, the time values may not be symmetric, i.e., the time to go directly from location i to j may not be the same as the time to go directly from location j to i. An input value of n = 0 will terminate input.

Output

For each test case, you should output a single number indicating the minimum time to deliver all of the pizzas and return to the pizzeria.

Sample Input

Sample Output

HINT

This problem is generally similar to the previous question : https://www.cnblogs.com/nlyzl/p/11307139.html

Must be read with from 0 ~ n, this problem requires a number from 0 point, so the initial f [1] [0] = 0; i.e. 0 0:00 clicked to the end number is 0;

i required from 1 ~ (1 << n + 1) -1;

j: n ~ 0 To reverse cycle, because the title does not say can not turn back, so we want n Backward ensure that currently go with the opposite of the values go all know, for DP min to take us.

J and k because we all start from zero, so no 1 << (j-1) and 1 << (k-1) when the left

Direct and 1 << j can be 1 << k;

Transfer: f [i] [j] = min (f [i] [j], min (f [i ^ 1 << j] [k], f [i] [k]) + dis [k] [j ]); // i.e., a state should start taking the current state min and then to add the distance, again transferred ( because they can turn back )

The last statistics we have + dis [i] [0]; // distance from any point to zero

code:

#include<bits/stdc++.h>
using namespace std;
int n;
int dis[21][21];
int ans;
int f[1<<20][21];
inline int read(){
     int x=0,f=1;
     char ch=getchar();
     while(ch<'0'||ch>'9'){
         if(ch=='-')
             f=-1;
         ch=getchar();
     }
     while(ch>='0'&&ch<='9'){
         x=(x<<1)+(x<<3)+(ch^48);
         ch=getchar();
     }
     return x*f;
}

inline void write(int x){
     char F[200];
     int tmp=x>0?x:-x ;
     if(x<0)putchar('-') ;
     int cnt=0 ;
        while(tmp>0)
        {
            F[cnt++]=tmp%10+'0';
            tmp/=10;
        }
        while(cnt>0)putchar(F[--cnt]) ;
}

int main()
{
    while(scanf("%d",&n)&&n!=0)
    {
        ans=INT_MAX;
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                dis[i][j]=read();
            }
        }
        memset(f,0x3f,sizeof(f));
        f[1][0]=0;
        for(int i=1;i<=(1<<n+1)-1;i++)
        {
            for(int j=n;j>=0;j--)
            {    
                
                if(i&1<<j)
                {
                    for(int k=0;k<=n;k++)
                    {
                        if(i&1<<k&&j!=k)
                        {
        f[i][j]=min(f[i][j],min(f[i^1<<j][k],f[i][k])+dis[k][j]);
                        }
                    }
                }
            }
        }
        for(int i=0;i<=n;i++)
            ans=min(ans,f[(1<<n+1)-1][i]+dis[i][0]);
        write(ans);
        cout<<endl;
    }
    return 0;    
}