luogu P2770 航空路线问题

本题可以把求解的目标转换成从1到N两条不相交的路径，回想上一题，通过拆点来限制一边只能过一次，capacity为1，cost为-1来跑最大费用流，注意1点和N点的capacity要为2，因为需要过2次，答案就是最大费用流-2，本题的收获是输出路径，从每个点的出点出发(虚点)枚举其连接的下一个入点(实点)，然后输出该入点，再dfs出点，如果只有一条边但有从1到N的边，也能输出答案，要考虑edge case

#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
typedef long long LL;

const int maxm = 1e5+5;
const int INF = 0x3f3f3f3f;

struct edge{
    int u, v, cap, flow, cost, nex;
} edges[maxm];

int head[maxm], cur[maxm], cnt, fa[maxm], d[maxm];
bool inq[maxm], vis[555];
map<string, int> cache;
map<int, string> store;

void init() {
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int cap, int cost) {
    edges[cnt] = edge{u, v, cap, 0, cost, head[u]};
    head[u] = cnt++;
}

void addedge(int u, int v, int cap, int cost) {
    add(u, v, cap, cost), add(v, u, 0, -cost);
}

bool spfa(int s, int t, int &flow, LL &cost) {
    //for(int i = 0; i <= n+1; ++i) d[i] = INF; //init()
    memset(d, 63, sizeof(d));
    memset(inq, false, sizeof(inq));
    d[s] = 0, inq[s] = true;
    fa[s] = -1, cur[s] = INF;
    queue<int> q;
    q.push(s);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        inq[u] = false;
        for(int i = head[u]; i != -1; i = edges[i].nex) {
            edge& now = edges[i];
            int v = now.v;
            if(now.cap > now.flow && d[v] > d[u] + now.cost) {
                d[v] = d[u] + now.cost;
                fa[v] = i;
                cur[v] = min(cur[u], now.cap - now.flow);
                if(!inq[v]) {q.push(v); inq[v] = true;}
            }
        }
    }
    if(d[t] == INF) return false;
    flow += cur[t];
    cost += 1LL*d[t]*cur[t];
    for(int u = t; u != s; u = edges[fa[u]].u) {
        edges[fa[u]].flow += cur[t];
        edges[fa[u]^1].flow -= cur[t];
    }
    return true;
}

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

void dfs1(int x, int N) {
    cout << store[x-N] << "\n";
    vis[x] = true;
    for(int i = head[x]; i != -1; i = edges[i].nex) {
        int v = edges[i].v;
        if(edges[i].flow>0 && v <= N) {
            dfs1(v+N, N);
            return;
        }
    }
}

void dfs2(int x, int N) {
    vis[x] = true;
    for(int i = head[x]; i != -1; i = edges[i].nex) {
        int v = edges[i].v;
        if(edges[i].flow>0 && v <= N && !vis[v+N]) dfs2(v+N, N);
    }
    cout << store[x-N] << "\n";
}

void run_case() {
    init();
    int N, V;
    bool flag = false;
    string str;
    cin >> N >> V;
    for(int i = 1; i <= N; ++i) {
        cin >> str;
        cache[str] = i, store[i] = str;
    }
    for(int i = 0; i < V; ++i) {
        string t1, t2;
        cin >> t1 >> t2;
        int u = cache[t1], v = cache[t2];
        if(u>v) swap(u, v);
        if(u == 1 && v == N) flag = true;
        addedge(u+N, v, 1, 0);
    }
    int s = 0, t = (N<<1)+2;
    addedge(s, 1, 2, 0), addedge(N<<1, t, 2, 0);
    for(int i = 1; i <= N; ++i) {
        if(i == 1 || i == N) addedge(i, i+N, 2, -1);
        else addedge(i, i+N, 1, -1);
    }
    LL cost = 0;
    int ans = MincostMaxflow(s, t, cost);
    if(ans == 2) {
        cout << -2-cost << "\n";
        dfs1(1+N,N), dfs2(1+N,N);
    } else if(ans == 1 && flag) {
        cout << "2\n";
        cout << store[1] << "\n" << store[N] << "\n" << store[1] << "\n";
    } else {
        cout << "No Solution!\n";
    }
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    run_case();
    cout.flush();
    return 0;
}

 

来源：https://www.cnblogs.com/GRedComeT/p/12301309.html