题意：
A-Z&&a-z 表示 集结点
从A点出发经过 最短步数 走到下一个集结点（A的下一个集结点为B ，Z的下一个集结点为a） 的路上遇到金子（*）则能够捡走（一个点仅仅能捡一次）
求从A点出发走遍全部的的集结点 最多能捡多少金子
思路：先对于第 i 个集结点用BFS求出 对于每一个点从该集结点所需的步数 为D[ I ] [ t ]
对于随意一个金子若 两个相邻的集结点的最短步数=其到该金子的步数和 则建一条边（能够拿）
然后最大流求
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cmath>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
#include <time.h>;
#define cler(arr, val) memset(arr, val, sizeof(arr))
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define IN freopen ("in.txt" , "r" , stdin);
#define OUT freopen ("out.txt" , "w" , stdout);
typedef long long LL;
const int MAXN = 10014;
const int MAXM = 41001;
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
struct Edge
{
int to,next,cap,flow;
} edge[MAXM]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{
tol = 0;
memset(head,-1,sizeof (head));
}
void addedge (int u,int v,int w,int rw = 0)
{
edge[tol].to = v;
edge[tol].cap = w;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = rw;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
gap[0] = 1;
int front = 0, rear = 0;
dep[end] = 0;
Q[rear++] = end;
while(front != rear)
{
int u = Q[front++];
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i]. to;
if(dep[v] != -1)continue;
Q[rear++] = v;
dep[v] = dep[u] + 1;
gap[dep[v]]++;
}
}
}
int S[MAXN];
int sap(int start,int end, int N)
{
BFS(start,end);
memcpy(cur,head,sizeof(head));
int top = 0;
int u = start;
int ans = 0;
int i;
while(dep[start] < N)
{
if(u == end)
{
int Min = INF;
int inser;
for( i = 0; i < top; i++)
{
if(Min > edge[S[i]].cap - edge[S[i]].flow)
{
Min = edge[S[i]].cap - edge[S[i]].flow;
inser = i;
}
}
for( i = 0; i < top; i++)
{
edge[S[i]]. flow += Min;
edge[S[i]^1].flow -= Min;
}
ans += Min;
top = inser;
u = edge[S[top]^1].to;
continue;
}
bool flag = false;
int v;
for( i = cur[u]; i != -1; i = edge[i]. next)
{
v = edge[i]. to;
if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
{
flag = true;
cur[u] = i;
break;
}
}
if(flag)
{
S[top++] = cur[u];
u = v;
continue;
}
int Min = N;
for( i = head[u]; i != -1; i = edge[i].next)
{
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
{
Min = dep[edge[i].to];
cur[u] = i;
}
}
gap[dep[u]]--;
if(!gap[dep[u]]) return ans;
dep[u] = Min + 1;
gap[dep[u]]++;
if(u != start)u = edge[S[--top]^1].to;
}
return ans;
}
char s[103][103];
bool vis[103][103];
int goal[103*103];
int ral[103*103];
int d[53][101*101];
int n,m;
int xx[4]= {1,0,-1,0};
int yy[4]= {0,-1,0,1};
void bfs(int x)
{
memset(d[x],INF,sizeof(d[x]));
cler(vis,false);
queue<int>q;
int t=ral[x];
vis[t/m][t%m]=true;
d[x][t]=0;
q.push(t);
while(!q.empty())
{
t=q.front();
q.pop();
int nx=t/m,ny=t%m;
for(int i=0; i<4; i++)
{
int dx=nx+xx[i],dy=ny+yy[i];
if(dx>=0&&dx<n&&dy>=0&&dy<m&&!vis[dx][dy]&&s[dx][dy]!='#')
{
vis[dx][dy]=true;
d[x][dx*m+dy]=d[x][t]+1;
q.push(dx*m+dy);
}
}
}
}
int find(char c)
{
if('A'<=c&&c<='Z')
return c-'A';
if('a'<=c&&c<='z')
return c-'a'+26;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
while(~scanf("%d%d",&n,&m))
{
init();
cler(ral,0);
int tol1=0,tol2=0;
for(int i=0; i<n; i++)
{
scanf("%s",s[i]);
for(int j=0; j<m; j++)
{
if(isalpha(s[i][j]))
{
ral[find(s[i][j])]=i*m+j;
tol1++;
}
else if(s[i][j]=='*')
goal[tol2++]=i*m+j;
}
}
for(int i=0; i<tol1; i++)
bfs(i);
int flag=0;
for(int i=1; i<tol1; i++)
for(int j=0; j<tol2; j++)
{ if(d[i][goal[j]]+d[i-1][goal[j]]==d[i-1][ral[i]])
addedge(i,tol1+j,1);
if(d[i-1][ral[i]]==INF)
flag=1;
}
if(flag)
printf("-1\n");
else
{
for(int i=1;i<tol1;i++)
addedge(0,i,1);
for(int i=0;i<tol2;i++)
addedge(tol1+i,tol1+tol2,1);
printf("%d\n",sap(0,tol1+tol2,tol1+tol2+1));
}
}
return 0;
}