bzoj 1179 缩点+bfs

嘛,考虑,图的大小以及强连通分量的性质,

可以轻易推出本题做法

#include<bits/stdc++.h>
#define MAXN 500005
using namespace std;

int n,m,tot,tot2,dx,tx,maxl;
int s,p,ed;
int h[MAXN],h2[MAXN],low[MAXN],dfn[MAXN],belong[MAXN];
int a[MAXN],sum[MAXN],f[MAXN];
bool vis[MAXN];
stack<int>q;

struct node{
    int from,to,next;
}e[MAXN<<1],e2[MAXN];

void init(){
    tot = tot2 = dx = tx = maxl = 0;
    memset(h,-1,sizeof(h));
    memset(h2,-1,sizeof(h2));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(f,0,sizeof(f));
    memset(sum,0,sizeof(sum));
    memset(a,0,sizeof(a));
    memset(vis,false,sizeof(vis));
}

void add(int x,int y){
    tot++;
    e[tot].from = x;
    e[tot].to = y;
    e[tot].next = h[x];
    h[x] = tot;
}

void add2(int x,int y){
    tot2++;
    e2[tot2].from = x;
    e2[tot2].to = y;
    e2[tot2].next = h2[x];
    h2[x] = tot2;
}

int tarjan(int now){
    dx++;
    low[now] = dfn[now] = dx;
    q.push(now),vis[now] = true;
    for(int i = h[now];i != (-1);i = e[i].next){
        if(!dfn[e[i].to]){
            tarjan(e[i].to);
            low[now] = min(low[now] , low[e[i].to]);
        }
        else if(vis[e[i].to]){
            low[now] = min(low[now] , dfn[e[i].to]);
        }
    }
    if(dfn[now]==low[now]){
        int u;
        tx++;
        do{
            u = q.top();
            belong[u] = tx;
            sum[tx]+=a[u];
            q.pop();
            vis[u] = false;
        }while(u!=now);
    }
}

void solve(){
    queue<int>t;
    t.push(belong[s]);
    f[belong[s]]=sum[belong[s]];
    while(!t.empty()){
        int u = t.front();t.pop();
        for(int i = h2[u];i!=(-1);i=e2[i].next){
            if(f[e2[i].to]<f[u]+sum[e2[i].to]){
                f[e2[i].to]=f[u]+sum[e2[i].to];
                t.push(e2[i].to);
            }
        }
    }
}

int main(){
    while(scanf("%d%d",&n,&m)==2){
        init();
        for(int i=1;i<=m;i++){
            int x,y;
            scanf("%d%d",&x,&y);
            add(x,y);
        }
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        scanf("%d%d",&s,&p);
        tarjan(s);
        for(int i=1;i<=m;i++){
            if(belong[e[i].from] == belong[e[i].to])continue;
            add2(belong[e[i].from] , belong[e[i].to]);        
        }
        solve();
        for(int i=1;i<=p;i++){
            scanf("%d",&ed);
            maxl = max(maxl,f[belong[ed]]);
        }
        cout<<maxl<<endl;
    }
}