Luogu P2770 航空路线问题

题目链接 \(Click\) \(Here\)

本来想调剂心情没想到写了那么久，还被\(dreagonm\)神仙嘲讽不会传纸条，我真是太弱了\(QAQ\)（原因：最开始写最大费用最大流一直想消圈，最后发现自己完全是\(zz\)了）

这个题是最大费用最大流，避免正环的关键在于只从西向东连边。还有要注意题目中并没有说能任一点开始结束，所以必须是两条\(1->n\)的路线。

路径输出方法真的是学到了，看下面代码吧。还有注意只有\(1->n\)一条边的特判。

#include <bits/stdc++.h>
using namespace std;

const int N = 400010;
const int M = 4000010;
const int INF = 0x3f3f3f3f;

int cnt = -1, head[N];

struct edge {
	int nxt, to, f, w;
}e[M];

void add_edge (int from, int to, int flw, int val) {
	e[++cnt].nxt = head[from];
	e[cnt].to = to;
	e[cnt].f = flw;
	e[cnt].w = val;
	head[from] = cnt;
}

void add_len (int u, int v, int f, int w) {
	add_edge (u, v, f, +w);
	add_edge (v, u, 0, -w);
}

int n, m;

map <string, int> mp;

string s1, s2, str[110];

int inn (int x) {return n * 0 + x;}
int out (int x) {return n * 1 + x;}

queue <int> q;
int vis[N], dis[N], flow[N];
int pre_edge[N], pre_node[N], max_flow, max_cost;

bool spfa (int s, int t) {
	memset (vis, 0, sizeof (vis));
	memset (dis, -0x3f, sizeof (dis));
	memset (flow, 0x3f, sizeof (flow));
	dis[s] = 0; vis[s] = true; q.push (s);
	while (!q.empty ()) {
		int u = q.front (); q.pop ();
		for (int i = head[u]; ~i; i = e[i].nxt) {
			int v = e[i].to;
			if (dis[v] < dis[u] + e[i].w && e[i].f) {
				dis[v] = dis[u] + e[i].w;
				flow[v] = min (flow[u], e[i].f);
				pre_edge[v] = i;
				pre_node[v] = u;
				if (!vis[v]) {
					vis[v] = true;
					q.push (v);
				}
			}
		}
		vis[u] = false;
	}
	return dis[t] != dis[0];
}

void dfs1 (int x) {
    cout << str[x - n] << endl;//第一遍dfs正序输出
    vis[x] = 1;//不让第二次dfs再找到这个点
    for (int i = head[x]; ~i; i = e[i].nxt) {
        if (e[i].to <= n && !e[i].f) {
			dfs1 (e[i].to + n);
			break;
		}//第一次dfs只找一条路径，找到就break
    }
}

void dfs2 (int x) {
    vis[x - n] = 1;
    for (int i = head[x]; ~i; i = e[i].nxt) {
        if (e[i].to <= n && !e[i].f && !vis[e[i].to + n]) {
			dfs2 (e[i].to + n);
		}//不走第一次路径走过的点
    }
    cout << str[x - n] << endl;//第二次dfs倒序输出
}//vis[n]在第一次dfs已经设为1，不会输出第二次

int main () {
	memset (head, -1, sizeof (head));
	cin >> n >> m;
	int s = n * 2 + 1;
	int t = n * 2 + 2;
	for (int i = 1; i <= n; ++i) {
		cin >> str[i]; mp[str[i]] = i;
		add_len (inn (i), out (i), 1, 1);
	}
	add_len (inn (1), out (1), 1, 0);
	add_len (inn (n), out (n), 1, 0);
	add_len (s, inn (1), 2, 0);
	add_len (out (n), t, 2, 0);
	bool have = false;
	for (int i = 1; i <= m; ++i) {
		cin >> s1 >> s2;
		if (mp[s1] > mp[s2]) swap (s1, s2);
		have |= (mp[s1] == 1 && mp[s2] == n);
    	add_len (out (mp[s1]), inn (mp[s2]), 1, 0);
	}
    max_flow = 0, max_cost = 0;
	while (spfa (s, t)) {
		max_flow += flow[t];
		max_cost += dis[t] * flow[t];
		int u = t;
		while (u != s) {
			e[pre_edge[u] ^ 0].f -= flow[t];
			e[pre_edge[u] ^ 1].f += flow[t];
			u = pre_node[u];
		}
	}
	if (max_flow == 1 && have) {
		cout << max_cost << endl;
		cout << str[1] << endl << str[n] << endl << str[1] << endl;
	} else if (max_flow == 2){
		memset (vis, 0, sizeof (vis));
		cout << max_cost << endl;
		dfs1 (n + 1); dfs2 (n + 1);
	} else puts ("No Solution!");
}