BZOJ 1179 Atm(强连通分量缩点+DP)

题目说可以通过一条边多次，且点权是非负的，所以如果走到图中的一个强连通分量，那么一定可以拿完这个强连通分量上的money。

所以缩点已经很明显了。缩完点之后图就是一个DAG，对于DAG可以用DP来求出到达每一个点的money最大值。具体实现我用的是bfs。

然后如果一个强连通分量内有酒馆，那么这个点就可以更新答案啦。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-3
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())==‘-‘) flag=1;
    else if(ch>=‘0‘&&ch<=‘9‘) res=ch-‘0‘;
    while((ch=getchar())>=‘0‘&&ch<=‘9‘)  res=res*10+(ch-‘0‘);
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar(‘-‘); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+‘0‘);
}
const int N=500005;
//Code begin...

struct Edge{int p, next;}edge[N], edge1[N];
int head[N], head1[N], cnt=1, cnt1=1, node[N], ans=0, dis[N];
int Low[N], DFN[N], Stack[N], Belong[N], Index, top, scc, num[N];
bool Instack[N], isjiu[N], jiu[N];
queue<int>Q;

void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;}
void add_edge1(int u, int v){edge1[cnt1].p=v; edge1[cnt1].next=head1[u]; head1[u]=cnt1++;}
void Tarjan(int u)
{
    int v;
    Low[u]=DFN[u]=++Index; Stack[top++]=u; Instack[u]=true;
    for (int i=head[u]; i; i=edge[i].next) {
        int v=edge[i].p;
        if (!DFN[v]) {
            Tarjan(v);
            if (Low[u]>Low[v]) Low[u]=Low[v];
        }
        else if (Instack[v]&&Low[u]>DFN[v]) Low[u]=DFN[v];
    }
    if (Low[u]==DFN[u]) {
        scc++;
        do{
            v=Stack[--top]; Instack[v]=false; Belong[v]=scc;
            num[scc]+=node[v]; jiu[scc]|=isjiu[v];
        }while (v!=u);
    }
}
void solve(int n){
    mem(DFN,0); mem(Instack,0); mem(num,0);
    Index=scc=top=0;
    FOR(i,1,n) if (!DFN[i]) Tarjan(i);
}
int main ()
{
    int n, m, u, v, s, p;
    n=Scan(); m=Scan();
    while (m--) u=Scan(), v=Scan(), add_edge(u,v);
    FOR(i,1,n) node[i]=Scan();
    s=Scan(); p=Scan();
    FOR(i,1,p) u=Scan(), isjiu[u]=1;
    solve(n);
    FO(i,1,n) for (u=head[i]; u; u=edge[u].next) {
        v=edge[u].p;
        if (Belong[v]==Belong[i]) continue;
        add_edge1(Belong[i],Belong[v]);
    }
    Q.push(Belong[s]); dis[Belong[s]]=num[Belong[s]];
    while (!Q.empty()) {
        u=Q.front(); Q.pop();
        if (jiu[u]) ans=max(ans,dis[u]);
        for (int i=head1[u]; i; i=edge1[i].next) {
            v=edge1[i].p;
            if (dis[v]>=dis[u]+num[v]) continue;
            dis[v]=dis[u]+num[v];
            Q.push(v);
        }
    }
    printf("%d\n",ans);
    return 0;
}