The path

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Special Judge

Problem Description

You have a connected directed graph.Let
d(x)
be the length of the shortest path from 1
to x.Specially
d(1)=0.A
graph is good if there exist x
satisfy d(1)<d(2)<....d(x)>d(x+1)>...d(n).Now
you need to set the length of every edge satisfy that the graph is good.Specially,if
d(1)<d(2)<..d(n),the
graph is good too.

The length of one edge must ∈
[1,n]

It‘s guaranteed that there exists solution.

Input

There are multiple test cases. The first line of input contains an integer
T,
indicating the number of test cases. For each test case:

The first line contains two integers n and m,the number of vertexs and the number of edges.Next m lines contain two integers each,
ui
and vi
(1≤ui,vi≤n),
indicating there is a link between nodes ui
and vi
and the direction is from ui
to vi.

∑n≤3?105,∑m≤6?105

1≤n,m≤105

Output

For each test case,print
m
lines.The i-th line includes one integer:the length of edge from
ui
to vi

Sample Input

2
4 6
1 2
2 4
1 3
1 2
2 2
2 3
4 6
1 2
2 3
1 4
2 1
2 1
2 1

Sample Output

1
2
2
1
4
4
1
1
3
4
4
4

Author

SXYZ

Source

2015 Multi-University Training Contest 8

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5421 5420 5419 5418 5417

我们如果知道每一个点的d[i] 显然cost[i,j]=d[j]-d[i]

我们从起点开始，每次从首或尾加一个节点进来，同时保证连通性

注意不能只找一次，因为首序列与尾序列大小无关

1->n->2->n-1->...->x 反例

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
#define MAXM (100000+10)
#pragma comment(linker, "/STACK:102400000,102400000")
#define mp make_pair
#define fi first
#define se second
#define pb push_back
typedef long long ll;

int n,m;
int u[MAXM],v[MAXM],ans[MAXM];

vector<int> To[MAXN];

int id[MAXN];
bool b[MAXN];
int main()
{
//	freopen("F.in","r",stdin);

	int T; cin>>T;
	while(T--) {
		MEM(u) MEM(v) MEM(ans) 

		cin>>n>>m;

		For(i,n) To[i].clear();

		For(i,m) {
			scanf("%d%d",&u[i],&v[i]);
			To[u[i]].pb(v[i]);
		}

		MEM(id) int cnt=0;
		MEM(b) b[1]=1;
		int l=1,r=n;
		while(l<=r) {
			if (b[l]) {
				id[l]=++cnt;
				int sz=To[l].size();
				Rep(i,sz) b[To[l][i]]=1;
				++l;
			}
			else if (b[r]) {
				id[r]=++cnt;
				int sz=To[r].size();
				Rep(i,sz) b[To[r][i]]=1;
				--r;
			}

		}

		For(i,m)
		{
			ans[i]=id[v[i]]-id[u[i]];
			if (ans[i]<=0||ans[i]>n) ans[i]=n;
			printf("%d\n",ans[i]);
		}

	} 	

	return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。