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（*）则可以捡走（一个点只能捡一次）

求从A点出发走遍所有的的集结点 最多能捡多少金子

思路：先对于第 i 个集结点用BFS求出 对于每个点从该集结点所需的步数  为D[ I ] [ t ] 

对于任意一个金子若  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（可以拿）

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;
}

【网络流】 HDU 3468 Treasure Hunting