Treasure Hunting HDU - 3468

题意：

　　输入一个n行m列的图

　　每次按字母顺序走最短路， 从一个字母走到下一个字母的过程中，只能拿走一个金子，求走完当前图中所有的字母后能拿到的金子的最大值

解析：

　　bfs求最短路

　　对于一个金子如果 dis1[i] + dis2[i] == dis1[next] 那么就代表着这个金子 在这条最短路上 可以拿 那么从上一个 字母 到当前节点连一条边 权值为1

　　会了吧。。。

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define pd(a) printf("%d\n", a);
#define plld(a) printf("%lld\n", a);
#define pc(a) printf("%c\n", a);
#define ps(a) printf("%s\n", a);
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = , INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
int n, m, s, t;
int d[maxn], vis[maxn];
int head[maxn], cur[maxn], cnt, dis1[maxn], dis2[maxn];
char str[][];
int dis[][] = {{, }, {, -}, {, }, {-, }};
vector<int> g;

struct edge
{
    int u, v, c, next;
}Edge[maxn];

void add_(int u, int v, int c)
{
    Edge[cnt].u = u;
    Edge[cnt].v = v;
    Edge[cnt].c = c;
    Edge[cnt].next = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int c)
{
    add_(u, v, c);
    add_(v, u, );
}

struct node
{
    int alpha, idx;
    bool operator < (const node &a) const{
        return alpha < a.alpha;
    }
}Node[maxn];

void bfs1(int s)
{
    for(int i = ; i < maxn; i++) d[i] = INF;
    queue<int> Q;
    mem(vis, );
    Q.push(s);
    vis[s] = ;
    d[s] = ;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = ; i < ; i++)
        {
            int v = u + dis[i][] * m + dis[i][];
            if(u % m ==  && i ==  || (u - ) % m ==  && i == ) continue;
            if(v <  || v > n * m || vis[v] || str[v / m + ][v % m] == '#') continue;
            d[v] = d[u] + ;
            vis[v] = ;
            Q.push(v);
        }
    }
}

bool bfs2()
{
    mem(d, );
    queue<int> Q;
    Q.push();
    d[] = ;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = head[u]; i != -; i = Edge[i].next)
        {
            edge e = Edge[i];
            if(!d[e.v] && e.c > )
            {
                d[e.v] = d[u] + ;
                Q.push(e.v);
                if(e.v == t) return ;
            }
        }
    }
    return d[t] != ;
}

int dfs(int u, int cap)
{
    int ret = ;
    if(u == t || cap == )
        return cap;
    for(int &i = cur[u]; i != -; i = Edge[i].next)
    {
        edge e = Edge[i];
        if(d[e.v] == d[u] +  && e.c > )
        {
            int V = dfs(e.v, min(e.c, cap));
            Edge[i].c -= V;
            Edge[i ^ ].c += V;
            ret += V;
            cap -= V;
            if(cap == ) break;
        }
    }
    if(cap > ) d[u] = -;
    return ret;
}

int Dinic()
{
    mem(d, );
    int ans = ;
    while(bfs2())
    {
        memcpy(cur, head, sizeof(head));
        ans += dfs(, INF);
    }
    return ans;
}

int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        g.clear();
        t = ;
        mem(head, -);
        cnt = ;
        int ans = ;
        for(int i = ; i <= n; i++)
        {
            rs(str[i] + );
            for(int j = ; j <= m; j++)
            {
                if(str[i][j] >= 'A' && str[i][j] <= 'z')
                    Node[ans].alpha = str[i][j], Node[ans++].idx = (i - ) * m + j;
                if(str[i][j] == '*')
                    g.push_back((i - ) * m + j), add((i - ) * m + j + , , );
            }
        }

        sort(Node, Node + ans);
        int num = , flag = ;
        for(int i = ; i < ans - ; i++)
        {
            ++num;
            bfs1(Node[i].idx);
            memcpy(dis1, d, sizeof(d));
            if(dis1[Node[i + ].idx] == INF)
            {
                flag = ;
                break;
            }
            bfs1(Node[i + ].idx);
            memcpy(dis2, d, sizeof(d));
            for(int j = ; j < g.size(); j++)
                if(dis1[g[j]] + dis2[g[j]] == dis1[Node[i + ].idx])
                    add(num, g[j] + , );
        }
        if(flag == )
        {
            cout << "-1" << endl;
            continue;
        }
        for(int i = ; i <= ; i++)
            add(, i, );

        cout << Dinic() << endl;
    }

    return ;
}