hdu 5067 Harry And Dig Machine

Harry And Dig Machine

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 320    Accepted Submission(s): 99

Problem Description

  As we all know, Harry Porter learns magic at Hogwarts School. However, learning magical knowledge alone is insufficient to become a great magician. Sometimes, Harry also has to gain knowledge from other certain subjects, such as language, mathematics, English, and even algorithm. 
  Dumbledore, the headmaster of Hogwarts, is planning to construct a new teaching building in his school. The area he selects can be considered as an n*m grid, some (but no more than ten) cells of which might contain stones. We should remove the stones there in order to save place for the teaching building. However, the stones might be useful, so we just move them to the top-left cell. Taking it into account that Harry learned how to operate dig machine in Lanxiang School several years ago, Dumbledore decides to let him do this job and wants it done as quickly as possible. Harry needs one unit time to move his dig machine from one cell to the adjacent one. Yet skilled as he is, it takes no time for him to move stones into or out of the dig machine, which is big enough to carry infinite stones. Given Harry and his dig machine at the top-left cell in the beginning, if he wants to optimize his work, what is the minimal time Harry needs to finish it?

 

Input

They are sever test cases, you should process to the end of file.
For each test case, there are two integers n and m.(1≤n,m≤50).
The next n line, each line contains m integer. The j-th number of ith line a[i][j] means there are a[i][j] stones on the jth cell of the ith line.( 0≤a[i][j]≤100 , and no more than 10 of a[i][j] will be positive integer).

 

Output

For each test case, just output one line that contains an integer indicate the minimal time that Harry can finish his job.

 

Sample Input

3 3

0 0 0

0 100 0

0 0 0

2 2

1 1

1 1

 

Sample Output

4 4

 

Source

BestCoder Round #14

 

 思路：哈密顿回路问题，dp[i][j] 表示状态为 i 时，最后一个节点是 j 时最短距离

状态的 i 的意思是，把 i 化为二进制，如果第 j 个点在里面，第j 位就为 1

比赛的时候，左右不分，也是醉了。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<ctime>
#include<bitset>
#define LL long long
#define maxn 110
#define INF 0x3f3f3f3f
using namespace std;

int mat[maxn][maxn],dis[150][150];
int map1[maxn][maxn], dp[(1<<10)+10][15] ;
bool vi[50][50];

struct node
{
    int x,y;
}qe[100];
int main()
{
    int i,j,k;
    int tmp,T,x,ans;
    int n,m,u;
    while(scanf("%d%d",&n,&m) != EOF)
    {
       for( i = 1 ; i <= n ;i++)
         for( j = 1 ; j <= m;j++)
            scanf("%d",&mat[i][j]) ;
       x = 0 ;
       for( i = 1 ; i <= n ;i++)
         for( j = 1 ; j <= m ;j++)
       {
           if(mat[i][j]>0){
                qe[x].x = i ;
                qe[x].y = j ;
                map1[i][j]=x++;
           }
           else map1[i][j]=-1;
       }
       for( i = 1 ; i <= n ;i++)
         for( j = 1 ; j <= m ;j++)if(mat[i][j]>0)
             {
                 for( k =0 ; k < x ;k++){
                    dis[map1[i][j]][map1[qe[k].x][qe[k].y]] = abs(qe[k].x-i)+abs(qe[k].y-j);
                 }
             }
       memset(dp,INF,sizeof(dp)) ;
       
       for( i = 0 ; i < x;i++){
          dp[1<<i][i] = abs(1-qe[i].x)+abs(1-qe[i].y);
       }
        for( i = 1 ; i < (1<<x);i++)
         for( j = 0 ; j < x ;j++)if(dp[i][j] != INF)
        {
            for( u = 0 ; u < x ;u++)if((i&(1<<u))==0)
            {
                dp[i|(1<<u)][u] = min(dp[i|(1<<u)][u],dis[j][u]+dp[i][j]) ;
            }
        }
        int ans=INF ;
        for( i = 0 ; i < x ;i++)
        {
            ans = min(ans,dp[(1<<x)-1][i]+abs(1-qe[i].x)+abs(1-qe[i].y)) ;
        }
        if(ans==INF)ans=0;
        printf("%d
",ans);
    }
    return 0 ;
}