题目传送门：https://www.luogu.org/problemnew/show/P3627
题解：tarjan强连通分量缩点+bfsspfa
代码如下：
1 #include<cstdio> 2 #include<iostream> 3 #define MN 1000010 4 using namespace std; 5 int n,m,w[MN],bar[MN],S,P,ans,hd[MN],cnt,sz[MN],CN,dis[MN],s[MN]; 6 int dye[MN],q[MN],top,dfn[MN],low[MN],dfs_ct,re_ct,re_hd[MN]; 7 bool vis[MN]; 8 struct node{int to,nex,d;}e[MN],re_e[MN]; 9 void add(int x,int y){e[++cnt].to=y;e[cnt].nex=hd[x];hd[x]=cnt;} 10 void tarjan(int x){ 11 vis[q[++top]=x]=true; 12 dfn[x]=low[x]=++dfs_ct; 13 for(int i=hd[x];i;i=e[i].nex){ 14 if(!dfn[e[i].to]){ 15 tarjan(e[i].to); 16 low[x]=min(low[x],low[e[i].to]); 17 } 18 else if(vis[e[i].to]) 19 low[x]=min(low[x],dfn[e[i].to]); 20 } 21 if(dfn[x]==low[x]){ 22 vis[x]=false; 23 sz[dye[x]=++CN]+=w[x]; 24 while(x!=q[top]){ 25 sz[dye[q[top]]=CN]+=w[q[top]]; 26 vis[q[top--]]=false; 27 } 28 --top; 29 } 30 } 31 void rebuild(){ 32 for(int i=1;i<=n;++i) 33 for(int j=hd[i];j;j=e[j].nex) 34 if(dye[i]!=dye[e[j].to]&&dye[i]&&dye[e[j].to]){ 35 re_e[++re_ct].to=dye[e[j].to]; 36 re_e[re_ct].d=sz[dye[e[j].to]]; 37 re_e[re_ct].nex=re_hd[dye[i]]; 38 re_hd[dye[i]]=re_ct; 39 } 40 } 41 bool spfa(int u){ 42 int head=1,tail=1; s[1]=u; 43 while(head<=tail){ 44 u=s[head++]; 45 for(int i=re_hd[u];i;i=re_e[i].nex){ 46 int v=re_e[i].to; 47 if(re_e[i].d+dis[u]>dis[v]){ 48 dis[v]=re_e[i].d+dis[u]; 49 s[++tail]=v; 50 } 51 } 52 } 53 } 54 int main() 55 { 56 scanf("%d%d",&n,&m); 57 for(int i=1;i<=m;i++){ 58 int x,y; scanf("%d%d",&x,&y); 59 add(x,y); 60 } 61 for(int i=1;i<=n;i++) scanf("%d",&w[i]); 62 scanf("%d%d",&S,&P); 63 for(int i=1;i<=P;i++){ 64 int x; scanf("%d",&x); 65 bar[x]=true; 66 } 67 for(int i=1;i<=n;i++)if(!dye[i]) tarjan(i); 68 rebuild(); 69 dis[dye[S]]=sz[dye[S]]; 70 spfa(dye[S]); 71 for(int i=1;i<=n;i++) 72 if(bar[i])if(dis[dye[i]]>ans) ans=dis[dye[i]]; 73 printf("%d",ans); 74 return 0; 75 }