[洛谷P3627][APIO2009]抢掠计划

题目大意：给你一张$n(n\leqslant5\times10^5)$个点$m(m\leqslant5\times10^5)$条边的有向图，有点权，给你起点和一些可能的终点。问从起点开始，到任意一个终点经过的点权和的最大值是多少。

题解：先把有向图缩点，然后从起点跑最长路，对每个终点取个最大值即可

卡点：求最长路写了$dijkstra$，然后死活调不出。



C++ Code：

#include <algorithm>
#include <cstdio>
#include <cctype>
#include <queue>
namespace __IO {
	int ch;
	inline int read() {
		static int x;
		while (isspace(ch = getchar())) ;
		for (x = ch & 15; isdigit(ch = getchar()); ) x = x * 10 + (ch & 15);
		return x;
	}
}
using __IO::read;

const int maxn = 500010, maxm = 500010;
int n, m;
int w[maxn], W[maxn];

namespace Graph {
	int head[maxn], cnt;
	struct Edge {
		int from, to, nxt;
	} e[maxm];
	inline void addedge(int a, int b) {
		e[++cnt] = (Edge) { a, b, head[a] }; head[a] = cnt;
	}

	int DFN[maxn], low[maxn], idx;
	int S[maxn], top, bel[maxn], scc;
	bool inS[maxn];
	void tarjan(int u) {
		DFN[u] = low[u] = ++idx;
		inS[S[++top] = u] = true;
		int v;
		for (int i = head[u]; i; i = e[i].nxt) {
			v = e[i].to;
			if (!DFN[v]) {
				tarjan(v);
				low[u] = std::min(low[u], low[v]);
			} else if (inS[v]) low[u] = std::min(low[u], DFN[v]);
		}
		if (DFN[u] == low[u]) {
			++scc;
			do {
				inS[v = S[top--]] = false;
				W[bel[v] = scc] += w[v];
			} while (v != u);
		}
	}
}

int head[maxn], cnt;
struct Edge {
	int to, nxt;
} e[maxm];
inline void addedge(int a, int b) {
	e[++cnt] = (Edge) { b, head[a] }; head[a] = cnt;
}

using Graph::scc;
using Graph::bel;
std::queue<int> q;
int dis[maxn];
bool inq[maxn];
void SPFA(int S) {
	dis[S] = W[S];
	q.push(S);
	while (!q.empty()) {
		int u = q.front(); q.pop();
		inq[u] = false;
		for (int i = head[u]; i; i = e[i].nxt) {
			int v = e[i].to;
			if (dis[v] < dis[u] + W[v]) {
				dis[v] = dis[u] + W[v];
				q.push(v);
				inq[v] = true;
			}
		}
	}
}

int S, ans;
int main() {
	n = read(), m = read();
	for (int i = 0, a, b; i < m; ++i) {
		a = read(), b = read();
		Graph::addedge(a, b);
	}
	for (int i = 1; i <= n; ++i) w[i] = read();
	S = read();
	Graph::tarjan(S);
	for (int i = 1, u, v; i <= m; ++i) {
		u = bel[Graph::e[i].from], v = bel[Graph::e[i].to];
		if (u != v) addedge(u, v);
	}
	SPFA(bel[S]);
	for (int i = read(), x; i; --i) ans = std::max(ans, dis[bel[x = read()]]);
	printf("%d\n", ans);
	return 0;
}