2015 Multi-University Training Contest 8 hdu 5385 The path

The path

Time Limit: 2000ms

Memory Limit: 65536KB

This problem will be judged on HDU. Original ID: 5385
64-bit integer IO format: %I64d      Java class name: Main

Special Judge

 

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 xsatisfy 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

Source

2015 Multi-University Training Contest 8 1006

 

解题：贪心

 

左边从2开始，右边从n开始，每次选与之前标记过的点相连的未标记过得点，该点的d[i]为该点加入的时间。最后输出时，判断该点是否在最短路上，不在的话，输出n，在的话输出d[v] - d[u]。

 #include <bits/stdc++.h>
 using namespace std;
 const int maxn = ;
 struct arc {
     int u,v,next;
     arc(int x = ,int y = ,int z = -) {
         u = x;
         v = y;
         next = z;
     }
 } e[maxn];
 int head[maxn],p[maxn],d[maxn],tot;
 void add(int u,int v) {
     e[tot] = arc(u,v,head[u]);
     head[u] = tot++;
 }
 void update(int u) {
     for(int i = head[u]; ~i; i = e[i].next)
         if(!p[e[i].v]) p[e[i].v] = u;
 }
 int main() {
     int kase,n,m,u,v;
     scanf("%d",&kase);
     while(kase--) {
         memset(head,-,sizeof head);
         memset(p,,sizeof p);
         scanf("%d%d",&n,&m);
         for(int i = tot = d[] = ; i < m; ++i) {
             scanf("%d%d",&u,&v);
             add(u,v);
         }
         d[] = d[n] = ;
         p[] = -;
         int L = , R = n,ds = ;
         while(L <= R) {
             if(p[L]) {
                 update(L);
                 d[L++] = ds++;
             }
             if(p[R]) {
                 update(R);
                 d[R--] = ds++;
             }
         }
         for(int i = ; i < tot; ++i)
             printf("%d\n",p[e[i].v] == e[i].u?d[e[i].v] - d[e[i].u]:n);
     }
     return ;
 }