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POJ-2195

• 这题使用的是最小费用流的模板。
• 建模的时候我的方法出现错误，导致出现WA，根据网上的建图方法没错。
• 这里的建图方法是每次到相邻点的最大容量为INF，而花费为1，因为花费等于距离。但是需要增加一个源点和一个汇点，然后将每个人和源点相连，每个房子和汇点相连，容量都为1，费用都为0.

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#include<stack>
#include<cstdio>
using namespace std;
#define INF 0x3f3f3f3f
#define M(a, b) memset(a, b, sizeof(a))
const int N = 1e4 + 5;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
int n, m;
struct Edge {
    int from, to, cap, flow, cost;
};

struct MCMF {
    int n, m;
    vector<Edge> edges;
    vector<int> G[N];
    int d[N], inq[N], p[N], a[N];

    void init(int n) {
        this->n = n;
        for (int i = 0; i <= n; ++i) G[i].clear();
        edges.clear();
    }

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

    bool spfa(int s, int t, int &flow, int &cost) {
        M(inq, 0); M(d, INF);
        d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
        queue<int> q;
        q.push(s);
        while (!q.empty()) {
            int x = q.front(); q.pop();
            inq[x] = 0;
            for (int i = 0; i < G[x].size(); ++i) {
                Edge &e = edges[G[x][i]];
                if (d[e.to] > d[x] + e.cost && e.cap > e.flow) {
                    d[e.to] = d[x] + e.cost;
                    p[e.to] = G[x][i];
                    a[e.to] = min(a[x], e.cap-e.flow);
                    if (inq[e.to]) continue;
                    q.push(e.to); inq[e.to] = 1;
                }
            }
        }
        if (d[t] == INF) return false;
        flow += a[t];
        cost += d[t] * a[t];
        int u = t;
        while (u != s) {
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -= a[t];
            u = edges[p[u]].from;
        }
        return true;
    }

    int Mincost(int s, int t) {
        int flow = 0, cost = 0;
        while (spfa(s, t, flow, cost));
        return cost;
    }

}solver;
char str[205][205];

bool check(int x, int y) {
    if (x>0 && x<=n && y>0 && y<=m) return 1;
    return 0;
}
struct node{
    int x;
    int y;
    int num;
};
node men[N];
node hou[N];
char ma[N][N];
int main() {
    while (scanf("%d%d", &n, &m), n&&m) {
        solver.init(n*m+1);
        for (int i = 1; i <= n; ++i) {
            scanf("%s", str[i]+1);
            for (int j = 1; j <= m; ++j) {
                if (str[i][j]=='H') solver.AddEdge((i-1)*m+j, n*m+1, 1, 0);
                if (str[i][j]=='m') solver.AddEdge(0, (i-1)*m+j, 1, 0);
                for (int k = 0; k < 4; ++k) {
                    int nx = i+dx[k], ny = j+dy[k];
                    if (check(nx, ny)) solver.AddEdge((i-1)*m+j, (nx-1)*m+ny, INF, 1);
                }
            }
        }
        printf("%d\n", solver.Mincost(0, n*m+1));

        //--------------------------------version2
        // int n1=201;
        // solver.init(2*n1+1);
        // int ansh=n,k2=0;
        // int ansm=0,k1=0;
        // for(int i=0;i<n;i++){
        //     scanf("%s", ma[i]);
        //     for(int j=0;j<m;j++){
        //         if(ma[i][j]=='m'){//人
        //             ++ansm;
        //             men[k1].x=i;
        //             men[k1].y=j;
        //             men[k1++].num=ansm;
        //             solver.AddEdge(0,ansm,1,0);
        //         }else if(ma[i][j]=='H'){
        //             ++ansh;
        //             hou[k2].x=i;
        //             hou[k2].y=j;
        //             hou[k2++].num=ansh;
        //             solver.AddEdge(ansh,2*n1+1,1,0);
        //         }
        //     }
        // }
        // for(int i=0;i<k1;i++){
        //     for(int j=0;j<k2;j++){
        //         int disc=abs(men[i].x-hou[j].x)+abs(men[i].y-hou[j].y);
        //         solver.AddEdge(men[i].num,hou[j].num,1,disc);
        //     }
        // }
        // printf("%d\n", solver.Mincost(0, n1*2+1));
    }
    return 0;
}

POJ-2195(最小费用最大流+MCMF算法)

标签：容量   tac   string   algorithm   inf   根据   fine   for   print   

原文地址：https://www.cnblogs.com/GarrettWale/p/11437993.html