bzoj 1179[Apio2009]Atm (tarjan+spfa)

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int cnt, hh[510000], hhh[510000], stack[510000];
int dfn[510000], low[510000], num, ans, q, top, w[510000];
int father[510000], dd[510000], j, n, h, t, l[510000], z, x, y, m, s, p;
bool d[510000];
struct node
{
    int v, next;
};
node b[510000], ss[510000];
void add(int aa, int bb)//连边
{
    b[++cnt].v = bb;
    b[cnt].next = hh[aa];
    hh[aa] = cnt;
}
void addd(int aa, int bb)//连接新建图上的边
{
    ss[++cnt].v = bb;
    ss[cnt].next = hhh[aa];
    hhh[aa] = cnt;
}



void tarjan(int k)
{
    int i;
    dfn[k] = low[k] = ++num;
    stack[++top] = k;
    d[k] = true;
    for(i = hh[k]; i != 0; i = b[i].next)
    {
        int e = b[i].v;
        if(!dfn[e])
        {
            tarjan(e);
            low[k] = min(low[k], low[e]);
        }
        else if(d[e] == true)
        {
            low[k] = min(low[k], dfn[e]);
        }
    }
    if(dfn[k] == low[k])
    {
        d[k] = false;
        father[k] = k;
        while(stack[top] != k)
        {
            w[k] += w[stack[top]];
            father[stack[top]] = k;
            d[stack[top--]] = false;
        }
        top--;
    }

}
void rebuild()
{
    cnt = 0;
    int i;
    for(i = 1; i <= n; i++)
    {
        for(j = hh[i]; j != 0; j = b[j].next)
        {
            t = b[j].v;
            if(father[i] == father[t])continue;
            addd(father[i], father[t]);
        }
    }
}
void spfa()
{
    int i;
    q = s;
    memset(d, 0, sizeof(d));
    h = 0, t = 0;
    l[q] = w[q];
    d[q] = true;
    while(1)
    {
        if(h > t)break;

        for(i = hhh[q]; i != 0; i = ss[i].next)
        {
            z = ss[i].v;
            if(l[q] + w[z] > l[z])
            {
                l[z] = l[q] + w[z];
                if(d[z])continue;
                dd[++t] = z;
                d[z] = true;
            }
        }
        d[q] = false;
        q = dd[++h];
    }
}
int main()
{
    int i;
    scanf("%d %d", &n, &m);
    for(i = 1; i <= m; i++)
    {
        scanf("%d %d", &x, &y);
        add(x, y);
    }
    for(i = 1; i <= n; i++)
    {
        scanf("%d", &w[i]);
    }
    scanf("%d %d", &s, &p);
    for(j = 1; j <= n; j++)
    {
        if(!dfn[j])tarjan(j);
    }
    rebuild();//利用tarjan求出所有强连通分量

    s = father[s];//如果起点在一个强连通分量中，那么将起点换成该强连通分量的根
                  //因为强连通分量上的点只要能到达一个就可以到达该强连通分量上的其它点，并且一条路可以走很多遍。
                  //重要的事说三遍

    spfa();//利用spfa求出最长路

    for(i = 1; i <= p; i++)
    {
        scanf("%d", &q);
        ans = max(ans, l[father[q]]);
    }
    printf("%d", ans);
    return 0;
}