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?

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).

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

3 3
0 0 0
0 100 0
0 0 0
2 2
1 1
1 1

4
4

heyang   |   We have carefully selected several similar problems for you:  5069 5068 5066 5065 5064

TSP问题，状压DP~~~~

设DP[i | (1 << j)][j] 表示当前走过的城市的状态是 i | (1 << j），当前处在城市j上的最短路径；

则 DP[i | (1 << j)][j] = min(DP [i | (1 << j)][j] , DP[i][k] + dist[k][j])

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <vector>  
#include <queue>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  

using namespace std;

int dp[2050][15];
int dist[15][15];
struct node
{
    int x, y;
}pp[15];

int main()
{
    int n, m;
    while (~scanf("%d%d", &n, &m))
    {
        int cnt = 1;
        int t;
        pp[0].x = 0;
        pp[0].y = 0;
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < m; ++j)
            {
                scanf("%d", &t);
                if (t)
                {
                    pp[cnt].x = i;
                    pp[cnt++].y = j;
                }
            }
        }
        for (int i = 0; i < cnt; ++i)
        {
            for (int j = i; j < cnt; ++j)
            {
                dist[i][j] = abs(pp[i].x - pp[j].x) + abs(pp[i].y - pp[j].y);
                dist[j][i] = dist[i][j];
            }
        }
        memset (dp, 0x3f3f3f3f, sizeof(dp) );
        dp[0][0] = 0;
        for (int i = 0; i < (1 << cnt); ++i)
        {
            for (int j = 0; j < cnt; ++j)
            {
                for (int k = 0; k < cnt; ++k)
                {
                    if(( i & (1 << j) ) == 0)
                    {
                        dp[i | (1 << j)][j] = min(dp[i | ( 1 << j)][j], dp[i][k] + dist[k][j]);
                    }
                }
            }
        }
        printf("%d\n", dp[(1 << cnt) - 1][0]);
    }
    return 0;
}