## Description

## Input

## Output

## Sample Input

3

30 -100 30

00000000

010203#0

00000000

## Sample Output

## HINT

50% Data of 1≤D≤3.

100% Data of 1≤D≤9,1≤N, M≤10,-10000≤Vi≤10000.

** This blog has detailed pictures and Analysis **

**http://blog.csdn.net/Phenix_2015/article/details/50739989**

** The problem is how to judge whether a bean is in a polygon .**

** There's actually a good way to judge , That's a horizontal ray , Look, there are several intersections with the polygon , There's an odd number of intersections in the polygon , Otherwise, outside the polygon .**

** So the state is compressed , For a state , The fewer steps, the better , And then it can be more based on the compressed state , Make the shortest path of the layered graph **

** Each step updates the current state **

** But there is another situation , Even number of intersections may also be surrounded .**

** As shown in the figure below ： **

** So in the process of transfer , Obviously, the horizontal movement does not affect the answer , Only vertical movement affects the position of the point **

** In special cases, each line segment is assumed to be open at the top and closed at the bottom , That is, only when the lower breakpoint intersects the ray can it be useful , Then the line segments in the same direction will only be counted once **

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

#include<queue>

using namespace std;

struct ZYYS

{

int x,y,s;

};

const int dx[]={,,,-};

const int dy[]={,-,,};

int d,px[],py[],n,m,val[],ans;

int dist[][][],inf;

bool vis[][][];

char ss[][];

int cross(int sx,int sy,int x,int y,int s)

{int i;

int p=max(sy,y);

for (i=;i<=d;i++)

if (py[i]==p&&px[i]<x)

s^=(<<i-);

return s;

}

void spfa(int sx,int sy)

{int i,j;

queue<ZYYS>Q;

memset(dist,/,sizeof(dist));

inf=dist[][][];

memset(vis,,sizeof(vis));

Q.push((ZYYS){sx,sy,});

dist[sx][sy][]=;

while (Q.empty()==)

{

ZYYS u=Q.front();

Q.pop();

vis[u.x][u.y][u.s]=;

for (i=;i<;i++)

{

int x=u.x+dx[i],y=u.y+dy[i];

if (x<||x>n||y<||y>m) continue;

if (ss[x][y]=='#'||ss[x][y]!='') continue;

int s=u.s;

if (i<=)

s=cross(u.x,u.y,x,y,s);

if (dist[x][y][s]>dist[u.x][u.y][u.s]+)

{

dist[x][y][s]=dist[u.x][u.y][u.s]+;

if (vis[x][y][s]==)

{

vis[x][y][s]=;

Q.push((ZYYS){x,y,s});

}

}

}

}

for (i=;i<=(<<d)-;i++)

{

int res=-dist[sx][sy][i];

for (j=;j<=d;j++)

if (i&(<<j-)) res+=val[j];

ans=max(ans,res);

}

}

int main()

{int i,j;

cin>>n>>m>>d;

for (i=;i<=d;i++)

{

scanf("%d",&val[i]);

}

for (i=;i<=n;i++)

{

scanf("%s",ss[i]+);

for (j=;j<=m;j++)

{

if (ss[i][j]>''&&ss[i][j]<='')

px[ss[i][j]-'']=i,py[ss[i][j]-'']=j;

}

}

for (i=;i<=n;i++)

{

for (j=;j<=m;j++)

if (ss[i][j]=='')

{

spfa(i,j);

}

}

cout<<ans;

}

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