网络流24题（11）航空路线问题（最大费用最大流）

思路：

1. 题目最终可以抽象为：最长不相交路径问题，两条从 s 到 t 的不相交路径的最大长度;

2. 把每个点分成 <i, X> <i, Y> 两点，引弧，容量为 1，费用为 1，特殊的：s, t 容量设置为 2，因为可以重复选择;

3. 对于 i, j 存在路径且 i < j 则 <i, Y> 向 <j, X> 引弧，容量为 1，费用为 0. 特殊的如果 i = s, j = t，容量设置为 2;

4. 求上述网络的最大费用最大流，把费用设置为负数，即可求最小费用最大流。结果 - 2 即是输出结果。

 

#include <iostream>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <algorithm>
using namespace std;

const int MAXN = 1010;
const int INFS = 0x3FFFFFFF;

struct edge {
    int from, to, cap, flow, cost;
    edge(int _from, int _to, int _cap, int _flow, int _cost)
        : from(_from), to(_to), cap(_cap), flow(_flow), cost(_cost) {}
};

class MCMF {
public:
    void initdata(int n) {
        this->n = n;
        edges.clear();
        for (int i = 0; i < n; i++)
            G[i].clear();
    }
    void addedge(int u, int v, int cap, int cost) {
        edges.push_back(edge(u, v, cap, 0, cost));
        edges.push_back(edge(v, u, 0, 0, -cost));
        G[u].push_back(edges.size() - 2);
        G[v].push_back(edges.size() - 1);
    }
    bool SPFA() {
        for (int i = 0; i < n; i++)
            d[i] = INFS, vis[i] = false;
        queue<int> Q;
        Q.push(s);
        d[s] = 0, vis[s] = true, p[s] = 0, a[s] = INFS;
        while (!Q.empty()) {
            int u = Q.front(); Q.pop();
            vis[u] = false;
            for (int i = 0; i < G[u].size(); i++) {
                edge& e = edges[G[u][i]];
                if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = G[u][i];
                    a[e.to] = min(a[u], e.cap - e.flow);
                    if (!vis[e.to]) { vis[e.to] = true; Q.push(e.to); }
                }
            }
        }
        return d[t] != INFS;
    }
    void augment(int& flow, int& cost) {
        flow += a[t];
        cost += a[t] * d[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;
        }
    }
    int mincost(int s, int t) {
        this->s = s, this->t = t;
        int flow = 0, cost = 0;
        while (SPFA()) {
            augment(flow, cost);
        }
        return cost;
    }
    bool isvalid(int delta) {
        if (edges[G[s][0]].flow == 2)
            return true;
        return false;
    }
    void getpath(int x, int delta, vector<int>& v) {
        for (int i = 0; i < G[x].size(); i++) {
            edge& e = edges[G[x][i]];
            if (e.flow == 1 && !vis[e.to]) {
                vis[e.to] = true;
                v.push_back(e.to);
                getpath(e.to + delta, delta, v);
                break;
            }
        }
    }
    void getans(int delta, vector<int>& v1, vector<int>& v2) {
        memset(vis, false, sizeof(vis));
        v1.push_back(s); 
        v2.push_back(s);
        getpath(s + delta, delta, v1);
        getpath(s + delta, delta, v2);
    }
private:
    vector<edge> edges;
    vector<int> G[MAXN];
    int n, s, t, d[MAXN], p[MAXN], a[MAXN];
    bool vis[MAXN];
};

MCMF mcmf;
string city[MAXN];
map<string, int> my;

int main() {
    int n, v;
    cin >> n >> v;
    int s = 1, t = n + n;
    mcmf.initdata(t + 1);
    for (int i = 1; i <= n; i++) {
        cin >> city[i];
        my[city[i]] = i;
        if (i == 1 || i == n)
            mcmf.addedge(i, i+n, 2, -1);
        else
            mcmf.addedge(i, i+n, 1, -1);
    }
    for (int i = 0; i < v; i++) {
        string a, b;
        cin >> a >> b;
        int k1 = min(my[a], my[b]);
        int k2 = max(my[a], my[b]);
        if (k1 == 1 && k2 == n)
            mcmf.addedge(k1+n, k2, 2, 0);
        else
            mcmf.addedge(k1+n, k2, 1, 0);
    }
    int cost = mcmf.mincost(s, t);
    if (mcmf.isvalid(n)) {
        cost = -cost-2;
        cout << cost << endl;
        vector<int> v1;
        vector<int> v2;
        mcmf.getans(n, v1, v2);
        for (int i = 0; i < v1.size(); i++)
            cout << city[v1[i]] << endl;
        for (int i = v2.size() - 1; i >= 0; i--)
            cout << city[v2[i]] << endl;
    } else {
        cout << "No Solution!" << endl;
    }
    return 0;
}