題目

思路

複習tarjan
我們維護兩個陣列dfn[]和low[]，dfn[u]表示頂點u第幾個被（首次）訪問，low[u]表示頂點u及其子樹中的點，通過非父子邊（回邊），能夠回溯到的最早的點（dfn最小）的dfn值（但不能通過連線u與其父節點的邊）。對於邊(u, v)，如果low[v]>=dfn[u]，此時u就是割點。

程式碼

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn = 10007;
const int maxm = 100007;  
struct node
{
    int x,y;
}E[maxm],ne[maxm];
int n,m,tm = 0,top = 0,tot = 0;
int W[maxn],dfn[maxn],low[maxn],Head[maxn],Next[maxm],S[maxn],V[maxn],sd[maxn];
int nh[maxn],nn[maxm],in[maxn],D[maxn];
void tarjan(int x)
{
    dfn[x] = low[x] = ++tm;
    S[++top] = x; V[x] = 1;
    for (int y,i = Head[x]; i; i = Next[i])
    {
        y = E[i].y;
        if (!dfn[y])
        {
            tarjan(y);
            low[x] = min(low[x],low[y]);
        }
        else if (V[y])
        {
            low[x] = min(low[x],low[y]);
        }
    }
    if (dfn[x] == low[x])
    {
        int y;
        while (y = S[top--])
        {
              V[y] = 0;
              sd[y] = x;
              if (y==x) break;
              W[x] +=W[y];
        }
    }
    return;
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i = 1; i<=n; i++)
    {
        scanf("%d",&W[i]);
    }
    for (int i = 1; i<=m; i++)
    {
        scanf("%d%d",&E[i].x,&E[i].y);
        Next[i] = Head[E[i].x];
        Head[E[i].x] = i;
    }    
    for (int i = 1; i<=n; i++)
    {
        if (!dfn[i])
        {
            tarjan(i);
        }
    }
    for (int x,y,i = 1; i<=m; i++)
    {
        x = sd[E[i].x]; y = sd[E[i].y];
        if (x!=y)
        {
            ne[++tot].x = x; ne[tot].y = y;
            nn[tot] = nh[x];
            nh[x] = tot;
            in[y]++;
        }
    }
    top =  0;
    for (int i = 1; i<=n; i++)
    {
        if (sd[i] == i && !in[i])
        {
            S[++top] = i;
            D[i] = W[i];
        }
    }
    while (top)
    {
        int x = S[top--];
        for (int y,i = nh[x]; i; i = nn[i])
        {
            y = ne[i].y;
            D[y] = max(D[y],D[x]+W[y]);
            if (--in[y] == 0)
            {
                S[++top] = y;
            }
        }
    }
    int ans = 0;
    for (int i = 1; i<=n; i++)
    {
        ans = max(ans,D[i]);
    }
    cout<<ans<<endl;
    return 0;
}