Luo Gu P4017 Topological sorting template

Los examples Valley P4017 : HTTPS: //www.luogu.com.cn/problem/P4017

 

Seeking the maximum number of the food chain, the biggest problem has been given in the definition of the food chain, the chain can be converted to a defined starting point is the point of 0 degree, a 0 degree point as the end point. Note the food web abstract of a directed graph may have more.

Idea: Just once topological sorting, dynamic cumulative record your answers to the final output to the maximum cumulative number to the end of the food chain.

template:

 1 bool toposort() {
 2     q = new queue();
 3     for (i = 0; i < n; i++)
 4         if (in_deg[i] == 0) q.push(i);
 5     ans = new vector();
 6     while (!q.empty()) {
 7         u = q.pop();
 8         ans.push_back(u);
 9         for each edge(u, v) {
10             if (--in_deg[v] == 0) q.push(v);
11         }
12     }
13     if (ans.size() == n) {
14         for (i = 0; i < n; i++)
15             std::cout << ans[i] << std::endl;
16         return true;
17     } else {
18         return false;
19     }
20 }

 

AC Code:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 5e3+5;
 5 const ll mod = 80112002;
 6 struct node{
 7     vector<int> v;
 8     int in,out;
 9 }g[maxn];
10 int n,m;
11 ll ans[maxn];
12 void add(int u,int v){
13     g[v].v.push_back(u);
14     g[v].out++,g[u].in++;   
15 }
16 void toposort(){
17     queue<int> q;
18     for(int i = 1;i<=n;i++) if(!g[i].in ) q.push(i),ans[i] = 1;
19     while(!q.empty()){
20         int cur = q.front();
21         q.pop();
22         for(int i = 0;i<g[cur].v.size();i++){
23             int u = g[cur].v[i];
24             ans[u] = (ans[u] + ans[cur])%mod;
25             if(--g[u].in == 0) q.push(u);  
26         }
27     }
28 }
29 int main(){
30     scanf("%d%d",&n,&m);
31     for(int i = 1;i<=m;i++){
32         int u,v;
33         scanf("%d%d",&u,&v);
34         add(u,v);
35     }
36     toposort();
37     ll res = 0;
38     for(int i = 1;i<=n;i++){
39         if(!g[i].out ) res = (res+ans[i])%mod;
40     }
41     cout<<res;
42     return 0;
43 }