感觉这两天怎么老是遇到状压啊。。。。

数字20以下，首想状压啊、、、

不过这题犯抽忘记考虑没有石头的时候了啊。

简单的状压：表示状态为j时以第i的作为结束。

PS：这题也在表扬大蓝翔的挖掘机技术啊。醉了啊。。。

Harry And Dig Machine

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL __int64
//#define LL long long
#define INF 0x3f3f3f
#define PI 3.1415926535898

const int maxn = 55;
int mp[maxn][maxn];
int n, m;
int dis[maxn][maxn];
using namespace std;
int dp[12][1<<12];

struct node
{
    int x, y;
} f[12];

int main()
{
    while(~scanf("%d %d", &n, &m))
    {
        int ans = 0;
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                scanf("%d",&mp[i][j]);
                if(mp[i][j])
                {
                    f[ans].x = i;
                    f[ans++].y = j;
                }
            }
        }
        memset(dis, 0, sizeof(dis));
        for(int i = 0; i < ans; i++)
            for(int j = 0; j < ans; j++) dis[i][j] = (abs(f[i].x-f[j].x)+abs(f[i].y-f[j].y));
        for(int i = 0; i < ans; i++)
            for(int j = 0; j < (1<<ans); j++) dp[i][j] = INF;
        for(int i = 0; i < ans; i++) dp[i][(1<<i)] = (abs(f[i].x-1)+abs(f[i].y-1));
        for(int j = 0; j < (1<<ans); j++)
        {
            for(int k = 0; k < ans; k++)
            {
                if(!(j&(1<<k))) continue;
                for(int p = 0; p < ans; p++)
                {
                    if(j&(1<<p)) continue;
                    dp[p][j|(1<<p)] = min(dp[k][j] + dis[k][p], dp[p][j|(1<<p)]);
                }
            }
        }
        int Min = INF;
        for(int i = 0; i < ans; i++) Min = min(Min, dp[i][(1<<ans)-1]+abs(f[i].x-1)+abs(f[i].y-1));
        if(Min == INF) Min = 0;
        printf("%d\n",Min);
    }
    return 0;
}