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+7;
const int M=1e4+7;
const int inf=0x3f3f3f3f;
const LL INF=0x3f3f3f3f3f3f3f3f;
const int mod=1e9 + 7;

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(1) {
            int now = st[--top];
            in[now] = false;
            id[now] = cnt;
            if(now == u) break;
        }
    }
}

int dp(int u) {
    if(f[u] != -1) return f[u];
    f[u] = 0;
    for(int i = 0; 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, -1, sizeof(head));
    memset(f, -1, sizeof(f));
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++) {
        int from, to; scanf("%d%d", &from, &to);
        add(from, to);
    }
    for(int i = 1; i <= n; i++)
        scanf("%d", &a[i]);

    scanf("%d%d", &S, &p);
    for(int i = 1; i <= p; i++) {
        int x; scanf("%d", &x);
        flag[x] = true;
    }
    for(int i = 1; i <= n; i++) {
        if(!dfn[i]) tarjan(i);
    }
    for(int i = 1; i <= n; i++) {
        sum[id[i]] += a[i];
        bar[id[i]] |= flag[i];
    }
    for(int i = 0; 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 = 0;
    for(int i = 1; i <= cnt; i++) {
        if(bar[i]) {
            ans = max(ans, dp(i));
        }
    }
    printf("%d\n", ans);
    return 0;
}
/*
*/