小w的铁路图「多校联考 2019」

注意

本文为愚蠢的暴力，仅能在奇怪的OJ上跑过奇怪的数据。

题意

给定一张有向图，对于每一条边，求删去该边后两端点最短路长度。

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思路

数据过水，直接bfs就过了。

代码

#include <bits/stdc++.h>

using namespace std;

namespace StandardIO {

	template<typename T>inline void read (T &x) {
		x=0;T f=1;char c=getchar();
		for (; c<'0'||c>'9'; c=getchar()) if (c=='-') f=-1;
		for (; c>='0'&&c<='9'; c=getchar()) x=x*10+c-'0';
		x*=f;
	}

	template<typename T>inline void write (T x) {
		if (x<0) putchar('-'),x*=-1;
		if (x>=10) write(x/10);
		putchar(x%10+'0');
	}

}

using namespace StandardIO;

namespace Project {
	
	const int N=1001;

	int n,m;
	int cnt;
	int head[N];
	struct node {
		int to,next;
	} edge[100001];
	int a[100001],b[100001];
	int vis[N],dis[N];
	queue<int> q;
	
	inline void add (int a,int b) {
		edge[++cnt].to=b,edge[cnt].next=head[a],head[a]=cnt;
	} 
	inline int bfs (int s,int t) {
		memset(vis,0,sizeof(vis));
		while (q.size()) q.pop();
		q.push(s),dis[s]=0,vis[s]=1;
		for (register int i=head[s]; i; i=edge[i].next) {
			int to=edge[i].to;
			if (to==t) continue;
			q.push(to),dis[to]=1,vis[to]=1;
		}
		q.pop();
		while (q.size()) {
			int now=q.front();q.pop();
			for (register int i=head[now]; i; i=edge[i].next) {
				int to=edge[i].to;
				if (vis[to]) continue;
				if (to==t) return dis[now]+1;
				dis[to]=dis[now]+1,vis[to]=1,q.push(to);
			}
		}
		return -1;
	}

	inline void MAIN () {
		read(n),read(m);
		for (register int i=1; i<=m; ++i) {
			read(a[i]),read(b[i]);
			add(a[i],b[i]);
		}
		for (register int i=1; i<=m; ++i) {
			write(bfs(a[i],b[i])),putchar(' ');
		}
	}
	
}

int main () {
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	Project::MAIN();
}