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P2770 航空路线问题

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P2770 航空路线问题

题面

题目链接

解法

$O(n^3)$的$dp$显然（bushi

考虑网络流的做法。首先每个点经过一次就想到拆点，然后按照原图连边，从$1$到$n$的最大流$\ge 2$就有解。

需要注意测试点$2$是$1$到$n$有连边且只有$1 \to n \to 1$这种走法的，但其实可以不用特判解决，输出路径时如果剩余流量$=+\infty-2$时多跑一次即可（具体见代码）。

AC代码

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/**
 * @file:           P2770.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P2770
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
    freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
    fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp &x) {
    bool sign = false;
    char ch = getchar();
    long double tmp = 1;
    for (; !isdigit(ch); ch = getchar())
        sign |= (ch == '-');
    for (x = 0; isdigit(ch); ch = getchar())
        x = x * 10 + (ch ^ 48);
    if (ch == '.')
        for (ch = getchar(); isdigit(ch); ch = getchar())
            tmp /= 10.0, x += tmp * (ch ^ 48);
    return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        write(x / 10);
    putchar((x % 10) ^ 48);
}
const int MAXN = 505;
const int MAXM = 2e4 + 10;
const int MAXS = 20;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct Dinic {
    struct Edge {
        int v, flow, cost;
    } edge[MAXM];
    int tot = 1, flow = 0, cost = 0;
    int head[MAXN], nxt[MAXM], dis[MAXN], cur[MAXN];
    bool vis[MAXN];
    void addedge(int u, int v, int flow, int cost) {
        edge[++tot] = {v, flow, cost};
        nxt[tot] = head[u], head[u] = tot;
        edge[++tot] = {u, 0, -cost};
        nxt[tot] = head[v], head[v] = tot;
    }
    bool spfa(int s, int t) {
        fill(vis, vis + MAXN, 0);
        fill(dis, dis + MAXN, -INF);
        queue<int> que;
        que.push(s);
        dis[s] = 0, vis[s] = 1;
        while (!que.empty()) {
            int u = que.front();
            que.pop(), vis[u] = 0;
            for (int i = head[u]; i; i = nxt[i]) {
                int v = edge[i].v;
                if (edge[i].flow && dis[v] < dis[u] + edge[i].cost) {
                    dis[v] = dis[u] + edge[i].cost;
                    if (!vis[v])
                        que.push(v), vis[v] = 1;
                }
            }
        }
        return dis[t] != -INF;
    }
    int augment(int u, int t, int mx) {
        if (u == t || mx == 0)
            return mx;
        vis[u] = 1;
        int ret = 0;
        for (int &i = cur[u]; i; i = nxt[i]) {
            int v = edge[i].v;
            if (vis[v] || dis[v] != dis[u] + edge[i].cost)
                continue;
            int tmp = augment(v, t, min(mx, edge[i].flow));
            cost += tmp * edge[i].cost;
            mx -= tmp, ret += tmp;
            edge[i].flow -= tmp, edge[i ^ 1].flow += tmp;
            if (mx == 0)
                break;
        }
        vis[u] = 0;
        return ret;
    }
    int mcmf(int s, int t) {
        while (spfa(s, t)) {
            copy(head, head + MAXN, cur);
            flow += augment(s, t, INF);
        }
        return flow;
    }
};
int n, m, s, t, num, pt[MAXN][2];
char buf[MAXS], str[MAXN][MAXS];
Dinic network;
void print(int u, int on) {
    if (u == n) {
        if (on == 1)
            printf("%s\n", str[u]);
        return;
    }
    if (on == 1)
        printf("%s\n", str[u]);
    for (int i = network.head[pt[u][1]]; i; i = network.nxt[i])
        if (network.edge[i].flow == INF - 1)
            print((network.edge[i].v - 1) >> 1, on);
    if (on == -1)
        printf("%s\n", str[u]);
}
signed main() {
    read(n), read(m);
    s = ++num, t = ++num;
    for (int i = 1; i <= n; ++i) {
        scanf("%s", str[i]);
        pt[i][0] = ++num;
        pt[i][1] = ++num;
    }
    network.addedge(s, pt[1][1], 2, 0);
    network.addedge(pt[n][0], t, 2, 0);
    for (int i = 1; i <= n; ++i)
        network.addedge(pt[i][0], pt[i][1], 1, 0);
    for (int i = 1; i <= m; ++i) {
        int u, v;
        scanf("%s", buf);
        for (int j = 1; j <= n; ++j)
            if (strcmp(buf, str[j]) == 0)
                u = j;
        scanf("%s", buf);
        for (int j = 1; j <= n; ++j)
            if (strcmp(buf, str[j]) == 0)
                v = j;
        if (u > v)
            swap(u, v);
        network.addedge(pt[u][1], pt[v][0], INF, 1);
    }
    if (network.mcmf(s, t) != 2)
        return puts("No Solution!"), 0;
    write(network.cost), putchar('\n');
    for (int i = network.head[pt[1][1]], on = 1; i; i = network.nxt[i]) {
        if (network.edge[i].flow < INF - 2)
            continue;
        for (int j = network.edge[i].flow; j < INF; ++j, on = -on) {
            if (on == 1)
                printf("%s\n", str[1]);
            print((network.edge[i].v - 1) >> 1, on);
            if (on == -1)
                printf("%s\n", str[1]);
        }
    }
    return 0;
}