The path
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 282 Accepted Submission(s): 76
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
Source 2015 Multi-University Training Contest 8
每個點距離1-n,知道距離後,U-v的邊就是abs(dist[u]-dist[v])
是以從1,n兩個方向開始bfs
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<set>
using namespace std;
#define maxn 100007
int head[maxn];
int cnt;
struct Edge{
int v,u,next;
};
int dist[maxn];
Edge edge[maxn];
void addedge(int u,int v){
edge[cnt].v = v;
edge[cnt].u = u;
edge[cnt].next = head[u];
head[u] = cnt++;
}
int check[maxn];
int main(){
int t,n,m;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
cnt = 0;
for(int i = 0;i <= n; i++){
check[i] = 0;
head[i] = -1;
}
int u,v;
for(int i = 0;i < m; i++){
scanf("%d%d",&u,&v);
addedge(u,v);
}
check[1] = 1;
check[n] = 1;
int l = 1, r = n;
for(int i = 1;i <= n; i++){
if(check[l] == 0)
u = r--;
else u = l++;
dist[u] = i;
for(int j = head[u]; j != -1; j=edge[j].next){
v = edge[j].v;
check[v] = 1;
}
}
for(int i = 0;i < m; i++){
u = edge[i].u;
v = edge[i].v;
if(dist[u] < dist[v])
printf("%d\n",dist[v]-dist[u]);
else printf("%d\n",n);
}
}
return 0;
}
/*
22
4 3
1 2
1 4
4 3
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
*/