The path

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 724    Accepted Submission(s): 277
Special Judge


Problem Description

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

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

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 ​​

 


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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;
}