【CF241E】Flights

【CF241E】Flights

题面

洛谷

题解

对于原来的图，如果一条边不出现在(1)到(n)的路径上面，直接(ban)掉即可。

那么考虑一条边(u ightarrow v)，一定满足(1leq dis_v-dis_uleq 2)，其中(dis_u,dis_v)表示(1)到(u,v)的最短路。直接根据这个性质跑差分约束即可，一条边的答案即为(dis_v-dis_u)。

代码

#include <iostream>
#include <cstdio> 
#include <cstdlib> 
#include <cstring> 
#include <cmath> 
#include <algorithm> 
#include <vector>
#include <queue> 
using namespace std; 
inline int gi() { 
    register int data = 0, w = 1; 
    register char ch = 0; 
    while (!isdigit(ch) && ch != '-') ch = getchar(); 
    if (ch == '-') w = -1, ch = getchar(); 
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar(); 
    return w * data; 
} 
const int INF = 1e9; 
const int MAX_N = 1e3 + 5, MAX_M = 5e3 + 5; 
struct Edge { int u, v; } a[MAX_M]; 
struct Graph { int to, cost; } ; 
vector<Graph> G[MAX_N]; 
vector<int> E[MAX_N]; 
int N, M, vis[MAX_N]; 
void bfs(int s, int op) { 
	queue<int> que; 
	que.push(s), ++vis[s]; 
	while (!que.empty()) { 
		int x = que.front(); que.pop(); 
		for (auto v : E[x]) 
			if (vis[v] == op) ++vis[v], que.push(v); 
	} 
} 
int dis[MAX_N]; 
bool inq[MAX_N]; 
bool spfa() { 
	static int cnt[MAX_N]; 
	queue<int> que; que.push(1), inq[1] = 1, ++cnt[1]; 
	for (int i = 2; i <= N; i++) dis[i] = INF; 
	while (!que.empty()) { 
		int x = que.front(); que.pop(); 
		for (auto e : G[x]) { 
			int v = e.to, w = e.cost; 
			if (dis[x] + w < dis[v]) { 
				dis[v] = dis[x] + w; 
				if (!inq[v]) ++cnt[v], inq[v] = 1, que.push(v); 
				if (cnt[v] >= N) return 0; 
			} 
		} 
		inq[x] = 0; 
	} 
	return 1; 
} 
int main () {
#ifndef ONLINE_JUDGE 
    freopen("cpp.in", "r", stdin); 
#endif 
	N = gi(), M = gi(); 
	for (int i = 1; i <= M; i++) { 
		a[i].u = gi(), a[i].v = gi(); 
		E[a[i].u].push_back(a[i].v); 
	} 
	bfs(1, 0); 
	for (int i = 1; i <= N; i++) E[i].clear(); 
	for (int i = 1; i <= M; i++) E[a[i].v].push_back(a[i].u); 
	bfs(N, 1); 
	for (int i = 1; i <= M; i++) { 
		int u = a[i].u, v = a[i].v; 
		if (vis[u] != 2 || vis[v] != 2) continue; 
		G[u].push_back((Graph){v, 2}); 
		G[v].push_back((Graph){u, -1}); 
	} 
	if (spfa()) puts("Yes"); 
	else return puts("No") & 0; 
	for (int i = 1; i <= M; i++) { 
		int u = a[i].u, v = a[i].v; 
		if (vis[u] != 2 || vis[v] != 2) puts("1"); 
		else printf("%d
", dis[v] - dis[u]); 
	} 
    return 0; 
}