Going Home POJ

就是一个简单题

四个月前a的一道题，今天又看到了，再a一遍吧。

好吧 我想多了 用了bfs求最短路  其实不用的 因为没有障碍物

#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
", a);
#define plld(a) printf("%lld
", a);
#define pc(a) printf("%c
", a);
#define ps(a) printf("%s
", 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 = 100100, INF = 0x7fffffff, LL_INF = 0x7fffffffffffffff;
int n, m, s, t, flow, value;
char str[110][110];
int head[maxn], d[maxn], cnt, vis[maxn], p[maxn], f[maxn], di[maxn];
int dis[4][2] = {{0, 1},{0, -1},{1, 0},{-1, 0}};
vector<int> g;
vector<int> ff;

void bfs(int s)
{
    mem(di, 0);
    queue<int> Q;
    Q.push(s);
    di[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = 0; i < 4; i++)
        {
            int v = u + dis[i][0] * m + dis[i][1];
            if(u % m == 0 && i == 0 || (u - 1) % m == 0 && i == 1) continue;
            if(v < 1 || v > n * m || di[v]) continue;
            di[v] = di[u] + 1;
            Q.push(v);
        }
    }
}

struct node
{
    int u, v, c, w, next;
}Node[maxn];

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

void add(int u, int v, int w, int c)
{
    add_(u, v, w, c);
    add_(v, u, -w, 0);
}

int spfa()
{
    for(int i = 0; i < maxn; i++) d[i] = INF;
    queue<int> Q;
    mem(vis, 0);
    mem(p, -1);
    Q.push(s);
    vis[s] = 1;
    d[s] = 0;
    p[s] = 0; f[s] = INF;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        vis[u] = 0;
      //  cout << u << endl;
        for(int i = head[u]; i != -1; i = Node[i].next)
        {
            node e = Node[i];
            //cout << e.v << endl;
            if(d[e.v] > d[u] + e.w && e.c > 0)
            {
                d[e.v] = d[u] + e.w;
                p[e.v] = i;
                f[e.v] = min(f[u], e.c);
                if(!vis[e.v])
                {
                    vis[e.v] = 1;
                    Q.push(e.v);
               //     cout << 11111 << endl;
                }
            }
        }
    }
    if(p[t] == -1) return 0;
    flow += f[t]; value += f[t] * d[t];
    for(int i = t; i != s; i = Node[p[i]].u)
    {
        Node[p[i]].c -= f[t];
        Node[p[i] ^ 1].c += f[t];
    }
    return 1;
}

void max_flow()
{
    flow = value = 0;
    while(spfa());
    printf("%d
", value);
}

int main()
{
    while(cin >> n >> m && n + m)
    {
        s = 0, t = n * m + 1;
        mem(head, -1);
        cnt = 0;
        g.clear(); ff.clear();
        for(int i = 1; i <= n; i++)
        {
            scanf("%s", str[i] + 1);
            for(int j = 1; j <= m; j++)
            {
                if(str[i][j] == 'm')
                    g.push_back((i - 1) * m + j), add(s, (i - 1) * m + j, 0, 1);
                else if(str[i][j] == 'H')
                    ff.push_back((i - 1) * m + j), add((i - 1) * m + j, t, 0, 1);
            }
        }
        for(int i = 0; i < g.size(); i++)
        {
            bfs(g[i]);
            for(int j = 0; j < ff.size(); j++)
            {
                add(g[i], ff[j], di[ff[j]] - 1, 1);
               // cout << g[i] << "  " << ff[j] << "  " << di[ff[j]] - 1 << endl;
            }
        }
        max_flow();

    }


    return 0;
}