题解【洛谷P3627】[APIO2009]抢掠计划

题面

首先比较容易发现一个性质：

如果可以进入一个强连通分量，那么这个强连通分量内的所有点就都可以被走到。

于是我们就可以考虑对整张图进行 Tarjan 缩点，每一个强连通分量的点权为这个强连通分量内部所有点的点权，这样整张图就变成了一个 DAG。

那么用 SPFA 跑一遍最长路，最后直接对 (dis[)每个酒吧所在的强连通分量编号(]) 取一个 (max) 即为答案。

注意实现上的一些细节。

#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)

using namespace std;

typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

inline int gi()
{
	int f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

inline LL gl()
{
	LL f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int INF = 0x3f3f3f3f, N = 1000003, M = N << 1;

int n, m;
int dfn[N], low[N], tim, stk[N], topp, scc_cnt, sz[N], id[N];
bool in_stk[N];
int tot, head[N], headc[N], ver[M], nxt[M], edge[N];
int a[N];
int S, P;
int bar[N];

inline void add(int u, int v)
{
	ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}

inline void add1(int u, int v, int w)
{
	ver[++tot] = v, edge[tot] = w, nxt[tot] = headc[u], headc[u] = tot;
}

void Tarjan(int u)
{
	dfn[u] = low[u] = ++tim, stk[++topp] = u, in_stk[u] = true;
	for (int i = head[u]; i; i = nxt[i])
	{
		int v = ver[i];
		if (!dfn[v])
		{
			Tarjan(v);
			low[u] = min(low[u], low[v]);
		}
		else if (in_stk[v]) low[u] = min(low[u], dfn[v]);
	}
	if (dfn[u] == low[u])
	{
		int y = -1;
		++scc_cnt;
		do
		{
			y = stk[topp--];
			in_stk[y] = false;
			sz[scc_cnt] += a[y];
			id[y] = scc_cnt;
		} while (y != u);
	}
}

int dis[N];
bool vis[N];

inline void SPFA(int s)
{
	dis[s] = sz[s];
	vis[s] = true;
	queue <int> q;
	q.push(s);
	while (!q.empty())
	{
		int u = q.front(); q.pop(); vis[u] = false;
		for (int i = headc[u]; i; i = nxt[i])
		{
			int v = ver[i], w = edge[i];
			if (dis[v] < dis[u] + w)
			{
				dis[v] = dis[u] + w;
				if (!vis[v])
					vis[v] = true, q.push(v);
			}
		}
	}
}

int main()
{
        n = gi(), m = gi();
	for (int i = 1; i <= m; i+=1)
	{
		int u = gi(), v = gi();
		add(u, v);
	}
	for (int i = 1; i <= n; i+=1) a[i] = gi();
	S = gi(), P = gi();
	for (int i = 1; i <= P; i+=1) bar[i] = gi();
	for (int i = 1; i <= n; i+=1) if (!dfn[i]) Tarjan(i);
	for (int u = 1; u <= n; u+=1)
		for (int i = head[u]; i; i = nxt[i])
		{
			int v = ver[i];
			if (id[u] != id[v])
				add1(id[u], id[v], sz[id[v]]);
		}
	SPFA(id[S]);
	int ans = 0;
	for (int i = 1; i <= P; i+=1) ans = max(ans, dis[id[bar[i]]]);
	printf("%d
", ans);
	return 0;
}