# Traversal of P3916 graph in Luogu

## Traversal of P3916 graph in Luogu

### Description

• The directed graph of N points and M edges is given. For each point v, find A(v) to represent the point with the largest number that can be reached from point v.

### Input

• Line 1, 2 integers N,M.

Next row M, each row has two integers Ui,Vi, representing the edge (Ui,Vi). Click 1, 2,..., N number.

### Output

• N integers A(1),A(2),... A(N)

4 3
1 2
2 4
4 3

4 4 3 4

### Explanation:

• When I wrote this question, I didn't see that the opposite side was built, but I saw that the second out was the shrinking point.
• Thinking of practicing hands, I made a reduced point template to cut this problem.
• The idea is to run DAG once after shrinking. Writing better with dfs than topology
```#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#define N 100005
using namespace std;

struct E {int next, to;} e[N];
int n, m, num;
int h[N], dp[N], u[N], v[N];

int dex, tot;
int low[N], dfn[N], obj[N], bel[N];
bool vis[N];
stack<int> stk;

{
int x = 0; char c = getchar();
while(c < '0' || c > '9') c = getchar();
while(c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return x;
}

{
e[++num].next = h[u];
e[num].to = v;
h[u] = num;
}

void tarjan(int x)
{
low[x] = dfn[x] = ++dex;
stk.push(x), vis[x] = 1;
for(int i = h[x]; i != 0; i = e[i].next)
{
int now = e[i].to;
if(!dfn[now])
tarjan(now),
low[x] = min(low[x], low[now]);
else if(vis[now])
low[x] = min(low[x], dfn[now]);
}
if(low[x] == dfn[x])
{
tot++;
while(1)
{
int now = stk.top();
stk.pop(), vis[now] = 0;
obj[tot] = max(obj[tot], now);
bel[now] = tot;
if(now == x) break;
}
}
}

void dfs(int x)
{
if(dp[x]) return;
dp[x] = obj[x];
for(int i = h[x]; i != 0; i = e[i].next)
{
dfs(e[i].to);
dp[x] = max(dp[x], dp[e[i].to]);
}
}

int main()
{
cin >> n >> m;
for(int i = 1; i <= m; i++)
{
}
for(int i = 1; i <= n; i++)
if(!dfn[i]) tarjan(i);
num = 0;
memset(h, 0, sizeof(h));
for(int i = 1; i <= m; i++)
if(bel[u[i]] != bel[v[i]])