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 = 301000;
struct arc {
    int u,v,next;
    arc(int x = 0,int y = 0,int z = -1) {
        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,-1,sizeof head);
        memset(p,0,sizeof p);
        scanf("%d%d",&n,&m);
        for(int i = tot = d[0] = 0; i < m; ++i) {
            scanf("%d%d",&u,&v);
            add(u,v);
        }
        d[1] = d[n] = 1;
        p[1] = -1;
        int L = 1, R = n,ds = 1;
        while(L <= R) {
            if(p[L]) {
                update(L);
                d[L++] = ds++;
            }
            if(p[R]) {
                update(R);
                d[R--] = ds++;
            }
        }
        for(int i = 0; i < tot; ++i)
            printf("%d
",p[e[i].v] == e[i].u?d[e[i].v] - d[e[i].u]:n);
    }
    return 0;
}