POJ-2195-Going Home(最小费用最大流)

链接:

https://vjudge.net/problem/POJ-2195

题意:

黄弘毅突然兴致大发准备免费送房子，m表示人，H表示房子，每个房子只能进一个人，房子数等于人数。黄弘毅为了方便起见决定要让所有人到自己的房子的距离和最小，请问这个距离和是多少？

思路:

最小费用最大流.
每个人连到每个房子.

代码:

#include <iostream>
#include <cstdio>
#include <vector>
#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;

const int MAXN = 1e2+10;
const int INF = 1e9;

struct Edge
{
    int from, to, flow, cap, cost;
    Edge(int f, int t, int flo, int ca, int c):from(f), to(t), flow(flo), cap(ca), cost(c){}
};

vector<Edge> edges;
vector<int> G[MAXN*2];
pair<int, int > H[MAXN], P[MAXN];
int Dis[MAXN*2], Vis[MAXN*2], Pre[MAXN*2];
int n, m, s, t;

void AddEdge(int from, int to, int cap, int cost)
{
    edges.push_back(Edge(from, to, 0, cap, cost));
    edges.push_back(Edge(to, from, 0, 0, -cost));
    G[from].push_back(edges.size()-2);
    G[to].push_back(edges.size()-1);
}

int GetLen(int a, int b)
{
    return abs(P[a].first-H[b].first)+abs(P[a].second-H[b].second);
}

bool SPFA()
{
    //SPFA求最短路,处理负权
    memset(Dis, MINF, sizeof(Dis));
    memset(Vis, 0, sizeof(Vis));
    queue<int> que;
    Dis[s] = 0;
    Vis[s] = 1;
    que.push(s);
    while (!que.empty())
    {
        int u = que.front();
        que.pop();
        Vis[u] = 0;
        for (int i = 0;i < G[u].size();i++)
        {
            Edge &e = edges[G[u][i]];
            if (e.cap > e.flow && Dis[e.to] > Dis[u]+e.cost)
            {
                Dis[e.to] = Dis[u]+e.cost;
                Pre[e.to] = G[u][i];
                if (!Vis[e.to])
                {
                    Vis[e.to] = 1;
                    que.push(e.to);
                }
            }
        }
    }
    if (Dis[t] != MINF)
        return true;
    return false;
}

int CostFlow()
{
    int cost = 0;
    while (SPFA())
    {
        int Min = INF;
        for (int i = t;i != s;i = edges[Pre[i]].from)
        {
            Edge &e = edges[Pre[i]];
            Min = min(Min, edges[Pre[i]].cap-edges[Pre[i]].flow);
        }
        for (int i = t;i != s;i = edges[Pre[i]].from)
        {
            edges[Pre[i]].flow += Min;
            edges[Pre[i]^1].flow -= Min;
        }
        cost += Dis[t];
    }
    return cost;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    while (cin >> n >> m && n)
    {
        int cntp = 0, cnth = 0;
        for (int i = s;i <= t;i++)
            G[i].clear();
        edges.clear();
        for (int i = 1;i <= n;i++)
        {
            for (int j = 1;j <= m;j++)
            {
                char op;
                cin >> op;
                if (op == 'H')
                    H[++cnth].first = i, H[cnth].second = j;
                if (op == 'm')
                    P[++cntp].first = i, P[cntp].second = j;
            }
        }
        //people i*2-1, home i*2
        for (int i = 1;i <= cntp;i++)
        {
            for (int j = 1;j <= cnth;j++)
            {
//                cout << GetLen(i, j) << endl;
                AddEdge(i*2-1, j*2, 1, GetLen(i, j));
            }
        }
        for (int i = 1;i <= cntp;i++)
            AddEdge(0, i*2-1, 1, 0);
        for (int i = 1;i <= cnth;i++)
            AddEdge(i*2, 2*cnth+1, 1, 0);
        s = 0, t = 2*cnth+1;
        int res = CostFlow();
        cout << res << endl;
    }

    return 0;
}