2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛 H Skiing【拓扑排序，关键路径】

2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛  H Skiing

In this winter holiday, Bob has a plan for skiing at the mountain resort.

This ski resort has MM different ski paths and NN different flags situated at those turning points.

The ii-th path from the S_iS​i​​-th flag to the T_iT​i​​-th flag has length L_iL​i​​.

Each path must follow the principal of reduction of heights and the start point must be higher than the end point strictly.

An available ski trail would start from a flag, passing through several flags along the paths, and end at another flag.

Now, you should help Bob find the longest available ski trail in the ski resort.

Input Format

The first line contains an integer TT, indicating that there are TT cases.

In each test case, the first line contains two integers NN and MM where 0 < N \leq 100000<N≤10000 and 0 < M \leq 1000000<M≤100000as described above.

Each of the following MM lines contains three integers S_iS​i​​, T_iT​i​​, and L_i~(0 < L_i < 1000)L​i​​ (0<L​i​​<1000) describing a path in the ski resort.

Output Format

For each test case, ouput one integer representing the length of the longest ski trail.

样例输入

1
5 4
1 3 3
2 3 4
3 4 1
3 5 2

样例输出

6

http://blog.csdn.net/yo_bc/article/details/77917288：

给定一个有向无环图，求最长路径。
由于是有向无环图，所以最长路肯定是一个入度为0到出度为0的路径，拓扑排序在确定当前点之前能够考虑到所有到它的情况，所以最后取个最值即可。
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 using namespace std;
 7 const int maxn=1e4+10;
 8 vector<pair<int,int> >edge[maxn];
 9 int T,n,m;
10 int deg[maxn],val[maxn];
11 queue<int> q;
12 
13 void init()
14 {
15     for(int i=0;i<=n;i++){
16         edge[i].clear();
17     }
18     memset(val,0,sizeof(val));
19     memset(deg,0,sizeof(deg));
20 }
21 
22 void toposort()
23 {
24     while(!q.empty()) q.pop();
25     for(int i=0;i<=n;i++){
26         if(!deg[i]) q.push(i);
27     }
28     while(!q.empty())
29     {
30         int u=q.front();q.pop();
31         for(int i=0;i<edge[u].size();i++){
32             int v=edge[u][i].first;
33             val[v]=max(val[v],val[u]+edge[u][i].second);
34             if(--deg[v]==0) q.push(v);
35         }
36     }
37     printf("%d\n",*max_element(val+1,val+1+n));
38 }
39 
40 int main()
41 {
42     scanf("%d",&T);
43     while(T--)
44     {
45         scanf("%d%d",&n,&m);
46         init();
47         int u,v,w;
48         for(int i=0;i<m;i++){
49             scanf("%d%d%d",&u,&v,&w);
50             edge[u].push_back(make_pair(v,w));
51             deg[v]++;
52         }
53         toposort();
54     }
55     return 0;
56 }

 






posted @ 2017-09-13 16:42  ╰追憶似水年華ぃ╮  阅读(282)  评论(0)  编辑  收藏  举报