「CF241E」Flights

传送门
Luogu

解题思路

首先对于所有不属于任何一条路径上的边，它的权值是任意的。
对于所有在路径上的边 ((u,v)) 满足 (1le dis_v-dis_ule2)
差分约束即可。

细节注意事项

• 用dfs判负环时注意一下时间效率

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <vector>
#include <cmath>
#include <ctime>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
	s = 0; int f = 0; char c = getchar();
	while (!isdigit(c)) f |= (c == '-'), c = getchar();
	while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
	s = f ? -s : s;
}

const int _ = 1010;
const int __ = 5010 * 2 + 1010;

int n, m, vis[_], dis[_], exi[_];
struct edge{ int x, y; }g[__];
vector < int > G1[_], G2[_];
int tot, head[_], nxt[__], ver[__], w[__];
inline void Add_edge(int u, int v, int d)
{ nxt[++tot] = head[u], head[u] = tot, ver[tot] = v, w[tot] = d; }

inline int check(int i) { return vis[g[i].x] == 3 && vis[g[i].y] == 3; }

inline int spfa(int u) {
	exi[u] = 1;
	for (rg int i = head[u]; i; i = nxt[i]) {
		int v = ver[i];
		if (dis[v] < dis[u] + w[i]) {
			dis[v] = dis[u] + w[i];
			if (exi[v]) return 0;
			if (!spfa(v)) return 0;
		}
	}
	exi[u] = 0;
	return 1;
}

inline void dfs1(int u) { vis[u] |= 1; for (rg int v : G1[u]) if (!(vis[v] & 1)) dfs1(v); }

inline void dfs2(int u) { vis[u] |= 2; for (rg int v : G2[u]) if (!(vis[v] & 2)) dfs2(v); }

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
#endif
	read(n), read(m);
	for (rg int u, v, i = 1; i <= m; ++i)
		read(u), read(v), G1[u].push_back(v), G2[v].push_back(u), g[i] = (edge) { u, v };
	dfs1(1), dfs2(n);
	for (rg int i = 1; i <= m; ++i)
		if (check(i)) Add_edge(g[i].x, g[i].y, 1), Add_edge(g[i].y, g[i].x, -2);
	for (rg int i = 1; i <= n; ++i) Add_edge(0, i, 0), dis[i] = -0x3f3f3f3f;
	if (!spfa(0)) return puts("No"), 0;
	puts("Yes");
	for (rg int i = 1; i <= m; ++i) {
		if (check(i)) printf("%d
", dis[g[i].y] - dis[g[i].x]);
		else puts("1");
	}
	return 0;
}

完结撒花 (qwq)