[NOIpTG2014]尋找道路——[SPFA]

【題意分析】

這是道挺好的思維題，但是被我秒了

首先直覺告訴我們連反邊，從終點開始往前進行bfs擴展標記，把那些可以到達終點的邊全都找出來。

但題目說能到達終點的點，他們連向的點全部都能到達終點，那麼我們就找那些沒有標記的點，他們連向的點全部是不符合要求的

這樣篩選過之後剩下的全都是符合題意的點，跑一遍最短路就好了

Code：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
#define INF 2147400000
#define MAXN 300000
using namespace std;

struct fls {
	int to, next;
}edge[MAXN];

bool vis[MAXN], Vis[MAXN];
int head[MAXN], q[MAXN], dist[MAXN], n, m, S, T, cnt;

inline int read () {
	register int s = 0, w = 1;
	register char ch = getchar ();
	while (! isdigit (ch)) {if (ch == '-') w = -1; ch = getchar ();}
	while (isdigit (ch)) {s = (s << 3) + (s << 1) + (ch ^ 48); ch = getchar ();}
	return s * w;
}

inline void connect (int u, int v) {
	if (u == v) return;
	edge[++cnt].to = v, edge[cnt].next = head[u], head[u] = cnt;
}

void BFS () {
	int t = 1, w = 1; q[t] = T, vis[T] = 1;
	while (t <= w) {
		for (register int i = head[q[t]]; i; i = edge[i].next) {
			int v = edge[i].to; if (vis[v]) continue;
			q[++w] = v, vis[v] = 1;
		}
		t++;
	}
}

inline void del (int now) {
	for (register int i = head[now]; i; i = edge[i].next) {
		int v = edge[i].to; if (vis[v]) Vis[v] = 0;
	}
}

void SPFA () {
	memset (vis, 0, sizeof vis); memset (q, 0, sizeof q);
	for (register int i = 1; i <= n; i++) dist[i] = INF;
	int t = 1, w = 1; q[t] = T, vis[T] = 1, dist[T] = 0;
	while (t <= w) {
		int now = q[t]; vis[now] = 1;
		for (register int i = head[now]; i; i = edge[i].next) {
			int v = edge[i].to; if (! Vis[v]) continue;
			if (dist[v] > dist[now] + 1) {
				dist[v] = dist[now] + 1;
				if (! vis[v]) vis[v] = 1, q[++w] = v;
			}
		}
		t++;
	}
}

int main () {
	n = read (), m = read ();
	for (register int i = 1; i <= m; i++) connect (read (), read ());
	S = read (), T = read (), BFS (), memcpy (Vis, vis, sizeof vis);
	for (register int i = 1; i <= n; i++)
		if (! vis[i]) del (i);
	SPFA (); return printf ("%d\n", dist[S] == INF ? -1 : dist[S]), 0;
}
/*
3 2  
1 2  
2 1  
1 3  
*/