题目链接：
题意：给出一幅图。图中有一些点，然后从第1个点出发，然后途径全部有石头的点。最后回到原点，然后求最小距离。当初作比赛的时候不知道这就是旅行商经典问题。回来学了一下。
思路:
状态转移方程
DP[k][i|base[k]]=min(DP[k][i|base[k]],DP[j][i]+dis[j][k])
DP[J][I]表示从起点到j点在i状态下的最小距离。。。DP[j][i]+dis[j][k]表从j到k的距离。。
。时间复杂度是(n∗m+(t2)∗(2t)），那么问题就得到了解决。。
题目：
Harry And Dig Machine
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 453 Accepted Submission(s): 164
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?
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 ofith 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, there are two integers n and m.
The next n line, each line contains m integer. The j-th number of
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
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#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=10+5;
int dp[maxn][1<<maxn],dis[maxn][maxn],base[maxn];//dp[j][i]表示在i状态下到达j的最小距离
int p[maxn][2],n,m;
int cal(int i,int j)
{
return abs(p[i][0]-p[j][0])+abs(p[i][1]-p[j][1]);
}
int main()
{
int cnt,k;
while(~scanf("%d%d",&n,&m))
{
cnt=0;
base[1]=1;
for(int i=2;i<=14;i++)
base[i]=base[i-1]<<1;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
scanf("%d",&k);
if(k||(i==1&&j==1))
{
p[++cnt][0]=i;
p[cnt][1]=j;
}
}
memset(dis,0,sizeof(dis));
for(int i=1;i<=cnt;i++)
for(int j=i+1;j<=cnt;j++)
dis[i][j]=dis[j][i]=cal(i,j);
memset(dp,INF,sizeof(dp));
dp[1][0]=0;
int lim=1<<cnt;
for(int i=0;i<lim;i++)//状态
for(int j=1;j<=cnt;j++)//j点为起点
{
if(dp[j][i]==INF) continue;
for(int k=1;k<=cnt;k++)//转移到的点
{
if(i&base[k]) continue;
dp[k][i|base[k]]=min(dp[k][i|base[k]],dp[j][i]+dis[j][k]);
}
}
printf("%d
",dp[1][lim-1]);
}
return 0;
}