A few days ago, Tom was tired of all the PC-games, so he went back to some old FC-games. "Hudson‘s Adventure Island" was his favorite which he had played thousands of times. But to his disappointed, the more he played, the more he lost. Tom soon gave up the game. 



Now, he invents an easier game on the base of “Hudson‘s Adventure Island”. He wants to know whether the new game is easy enough. So he came to you for help. 
To simplify the problem, you can assume there is matrix-map contains m(0 < m < 256) rows, n(0 < n < 256) columns, an entrance on the top-left corner (0, 0), an exit on the bottom-right corner (m-1, n-1). Each entry of the matrix contains a integer k.The range of k is defined below:
a)k = 0,it is a free space one can go through.
b)k = -1,it is an obstacle one can’t go through.
c)0 < k < 10000,it is a fruit one can go through and gain k points of energy.
d)k >= 0 at the entrance point and the end point.
At the begin, the hero has 0 points of energy and he can go four directions (up, down, left, right). Each move he makes cost 1 point of energy. No energy no move. If he can’t make a move or get to the exit, he loses the game. The number of fruit is 17 at most.

4 4
8 0 0 0
-1 -1 -1 0
-1 -1 -1 0
0 6 0 0
3 3
4 0 0
0 0 0
0 0 0
4 4
5 0 0 0
0 -1 -1 0
0 -1 -1 0
0 0 0 0

4
0
you loss!


Hint
The hero can pass the exit. When he gets the exit point, he can choose to get out to finish the game or move on to gain more points of energy.
 

思路：先用BFS 求出各个水果和终点之间的距离，再用状态压缩DP求解。

#include <cstdio>
#include <queue>
#define INF 999999999
#define max(A,B)(A>B?A:B)
using namespace std;

struct S{
int x,y;
}fruit[20],t;

int n,m,cnt,ans,mp[256][256],dis[20][20],cost[256][256],nxt[4][2]={{1,0},{0,1},{-1,0},{0,-1}},mx[1<<18][20],val[20];

void bfs(S t)
{
    int i,j,nx,ny;

    for(i=0;i<n;i++) for(j=0;j<m;j++) cost[i][j]=INF;

    queue<S>que;

    cost[t.x][t.y]=0;

    que.push(t);

    while(!que.empty())
    {
        t=que.front();

        for(i=0;i<4;i++)
        {
            nx=t.x+nxt[i][0];
            ny=t.y+nxt[i][1];

            if(nx>=0 && nx<n && ny>=0 && ny<m && mp[nx][ny]!=-1 && cost[nx][ny]==INF)
            {
                cost[nx][ny]=cost[t.x][t.y]+1;

                t.x=nx;
                t.y=ny;

                que.push(t);
            }

            t=que.front();
        }

        que.pop();
    }
}

int main()
{
    int i,j;

    while(~scanf("%d%d",&n,&m))
    {
        cnt=0;

        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                scanf("%d",&mp[i][j]);

                if(mp[i][j]>0)
                {
                    fruit[cnt].x=i;
                    fruit[cnt].y=j;

                    val[cnt++]=mp[i][j];
                }
            }
        }

        if(n==1 && m==1)
        {
            if(mp[0][0]>=0) printf("%d\n",mp[0][0]);
            else printf("you loss!\n");

            continue;
        }

        if(mp[0][0]<=0)
        {
            printf("you loss!\n");

            continue;
        }

        if(mp[n-1][m-1]==-1)
        {
            printf("you loss!\n");

            continue;
        }


        fruit[cnt].x=n-1;//终点也放进去
        fruit[cnt].y=m-1;


        for(i=0;i<=cnt;i++)
        {
            bfs(fruit[i]);

            for(j=0;j<=cnt;j++) dis[i][j]=cost[fruit[j].x][fruit[j].y];//各个水果和终点之间的距离

        }


        queue<S>que;

        for(i=1;i<(1<<cnt);i++) for(j=0;j<cnt;j++) mx[i][j]=-1;

        int v;

        t.x=1;
        t.y=0;

        mx[t.x][t.y]=val[0];

        que.push(t);

        while(!que.empty())
        {
            t=que.front();

            for(i=1;i<cnt;i++)
            {
                if(!(t.x & (1<<i)))
                {
                    v=mx[t.x][t.y]-dis[t.y][i];

                    if(v>=0)
                    {
                        t.x=(t.x | (1<<i));
                        t.y=i;

                        if(v+val[i]>mx[t.x][i])
                        {
                            mx[t.x][i]=v+val[i];
                            que.push(t);
                        }

                        t=que.front();
                    }
                }
            }

            que.pop();
        }

        ans=-1;

        for(i=1;i<(1<<cnt);i++) for(j=0;j<cnt;j++) ans=max(ans,mx[i][j]-dis[j][cnt]);

        if(ans>=0) printf("%d\n",ans);
        else puts("you loss!");
    }
}