bzoj 1179

题意：给你一个有向图，每个点有一个权值，有一个起点和q个终点，没经过一个点加上这个点的权值，让你选一条路，问你最大值是多少。

思路：tarjan强连通缩个点， 然后在拓扑图上dp一下就好啦， 注意第二次建图建反向边会好一点。

 #include<bits/stdc++.h>
 #define LL long long
 #define fi first
 #define se second
 #define mk make_pair
 #define pii pair<int,int>

 using namespace std;

 const int N=5e5+;
 const int M=1e4+;
 const int inf=0x3f3f3f3f;
 const LL INF=0x3f3f3f3f3f3f3f3f;
 const int mod=1e9 + ;

 struct edge {
     int from, to, nx;
 }e[N];
 bool bar[N], flag[N], in[N], ok[N];
 int n, m, tot, cnt, S, p, idx, top, head[N], a[N], sum[N], dfn[N], low[N], id[N], f[N], st[N];

 vector<int> edge[N];
 void add(int from, int to) {
     e[tot].from = from;
     e[tot].to = to;
     e[tot].nx = head[from];
     head[from] = tot++;
 }

 void tarjan(int u) {
     ++idx;
     dfn[u] = low[u] = idx;
     st[top++] = u; in[u] = true;
     for(int i = head[u]; ~i; i = e[i].nx) {
         int v = e[i].to;
         if(!dfn[v]) {
             tarjan(v);
             low[u] = min(low[u], low[v]);
         } else if(in[v]) {
             low[u] = min(low[u], dfn[v]);
         }
     }
     if(low[u] == dfn[u]) {
         cnt++;
         while() {
             int now = st[--top];
             in[now] = false;
             id[now] = cnt;
             if(now == u) break;
         }
     }
 }

 int dp(int u) {
     if(f[u] != -) return f[u];
     f[u] = ;
     for(int i = ; i < edge[u].size(); i++) {
         int v = edge[u][i];
         int ret = dp(v);
         if(ok[v]) {
             ok[u] = true;
             f[u] = max(f[u], f[v]);
         }
     }
     if(!ok[u]) return f[u];
     return f[u] = f[u] + sum[u];
 }
 int main() {
     memset(head, -, sizeof(head));
     memset(f, -, sizeof(f));
     scanf("%d%d", &n, &m);
     for(int i = ; i <= m; i++) {
         int from, to; scanf("%d%d", &from, &to);
         add(from, to);
     }
     for(int i = ; i <= n; i++)
         scanf("%d", &a[i]);

     scanf("%d%d", &S, &p);
     for(int i = ; i <= p; i++) {
         int x; scanf("%d", &x);
         flag[x] = true;
     }
     for(int i = ; i <= n; i++) {
         if(!dfn[i]) tarjan(i);
     }
     for(int i = ; i <= n; i++) {
         sum[id[i]] += a[i];
         bar[id[i]] |= flag[i];
     }
     for(int i = ; i < tot; i++) {
         int u = e[i].from, v = e[i].to;
         if(id[u] != id[v]) {
             edge[id[v]].push_back(id[u]);
         }
     }
     S = id[S];
     ok[S] = true;
     int ans = ;
     for(int i = ; i <= cnt; i++) {
         if(bar[i]) {
             ans = max(ans, dp(i));
         }
     }
     printf("%d\n", ans);
     return ;
 }
 /*
 */