[BZOJ1179] [Apio2009]Atm（tarjan缩点 + spfa）

传送门

题意

N个点M条边的有向图

每个点有点权

从某一个结点出发

问能获得的最大点权和

一个点的点权最多被计算一次

N<=500000 M<=500000

思路

先tarjan缩点，然后就形成一个dag，无环，所以直接spfa求最长路就行。

也可以先缩点，然后拓扑排序 + dp 搞。

代码

 #include <map>
 #include <queue>
 #include <stack>
 #include <cstdio>
 #include <cstring>
 #include <iostream>

 const int MAXN = ;
 int n, m, s, p, cnt, cnt1, tim, sz, ans;
 int head[MAXN], to[MAXN], next[MAXN], head1[MAXN], to1[MAXN], next1[MAXN];
 int a[MAXN], dfn[MAXN], low[MAXN], belong[MAXN], val[MAXN], dis[MAXN];
 bool ins[MAXN], vis[MAXN];
 std::map <int, int> map1[MAXN];
 std::stack <int> S;
 std::queue <int> q;

 inline int max(int x, int y)
 {
     return x > y ? x : y;
 }

 inline int min(int x, int y)
 {
     return x < y ? x : y;
 }

 inline int read()
 {
     int f = , x = ;
     char ch = getchar();
     for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -;
     for(; isdigit(ch); ch = getchar()) x = (x << ) + (x << ) + ch - '';
     return f * x;
 }

 inline void add(int x, int y)
 {
     to[cnt] = y;
     next[cnt] = head[x];
     head[x] = cnt++;
 }

 inline void add1(int x, int y)
 {
     to1[cnt1] = y;
     next1[cnt1] = head1[x];
     head1[x] = cnt1++;
 }

 inline void dfs(int u)
 {
     int i, v;
     S.push(u);
     ins[u] = ;
     dfn[u] = low[u] = ++tim;
     for(i = head[u]; i ^ -; i = next[i])
     {
         v = to[i];
         if(!dfn[v])
         {
             dfs(v);
             low[u] = min(low[u], low[v]);
         }
         else if(ins[v]) low[u] = min(low[u], dfn[v]);
     }
     if(!(low[u] ^ dfn[u]))
     {
         sz++;
         do
         {
             v = S.top();
             S.pop();
             belong[v] = sz;
             ins[v] = ;
         }
         while(u ^ v);
     }
 }

 inline void spfa()
 {
     int i, u, v;
     dis[belong[s]] = val[belong[s]];
     q.push(belong[s]);
     vis[belong[s]] = ;
     while(!q.empty())
     {
         u = q.front();
         q.pop();
         vis[u] = ;
         for(i = head1[u]; i ^ -; i = next1[i])
         {
             v = to1[i];
             if(dis[v] < dis[u] + val[v])
             {
                 dis[v] = dis[u] + val[v];
                 if(!vis[v])
                 {
                     vis[v] = ;
                     q.push(v);
                 }
             }
         }
     }
 }

 int main()
 {
     int i, x, y, u, v;
     n = read();
     m = read();
     memset(head, -, sizeof(head));
     memset(head1, -, sizeof(head1));
     for(i = ; i <= m; i++)
     {
         x = read();
         y = read();
         add(x, y);
     }
     for(i = ; i <= n; i++) a[i] = read();
     s = read();
     p = read();
     dfs(s);
     for(i = ; i <= n; i++) val[belong[i]] += a[i];
     for(u = ; u <= n; u++)
         for(i = head[u]; i ^ -; i = next[i])
         {
             v = to[i];
             if(belong[u] ^ belong[v] && !map1[belong[u]][belong[v]])
                 map1[belong[u]][belong[v]] = , add1(belong[u], belong[v]);
         }
     spfa();
     for(i = ; i <= p; i++)
     {
         x = read();
         ans = max(ans, dis[belong[x]]);
     }
     printf("%d\n", ans);
     return ;
 }