题意

给出一个图，上面有若干连通块。
对于每个连通块，为有向边组成的图，每个点上有点权。
最多在\(K+1\)个连通块操作，使得遍历这个连通块获得的点权之和最大。

思路

将每个连通块上的环缩点，变成一个DAG然后dp即可。

代码

#include <queue>
#include <stack>
#include <cstdio>
#include <algorithm>

const int MAXN = 200001, MAXM = 2000001;
std::stack<int> s;
std::queue<int> q;
int n, m, k, tot, tot1, dfncnt, col, ans;
int low[MAXN], dfn[MAXN], in_s[MAXN], scc[MAXN], ver[MAXM], next[MAXM], head[MAXM];
int ver1[MAXM], next1[MAXM], head1[MAXM], deg[MAXM];
int vw[MAXN], vv[MAXN], fa[MAXN], used[MAXN];
struct node {
	int id, a;
} f[MAXN];

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

void add1(int u, int v) {
	ver1[++tot1] = v;
	next1[tot1] = head1[u];
	head1[u] = tot1;	
}

void tarjan(int u) {
	low[u] = dfn[u] = ++dfncnt;
	s.push(u);
	in_s[u] = 1;
	int v;
	for (int i = head[u]; i; i = next[i])
		if (!dfn[v = ver[i]]) {
			tarjan(v);
			low[u] = std::min(low[u], low[v]);
		} else if (in_s[v])
			low[u] = std::min(low[u], dfn[v]);
	if (dfn[u] == low[u]) {
		++col;
		fa[col] = col;
		int p;
		while ((p = s.top()) != u) {
			vv[col] += vw[p];
			scc[p] = col;
			in_s[p] = 0;
			s.pop();
		}
		scc[p = s.top()] = col;
		vv[col] += vw[p];
		f[col].a = vv[col];
		f[col].id = col;
		in_s[p] = 0;
		s.pop();
	}
}

int find(int x) {
	return fa[x] = fa[x] == x ? x : find(fa[x]);
}

void dp() {
	for (int i = 1; i <= col; i++)
		if (!deg[i])
			q.push(i);
	while (q.size()) {
		int u = q.front(), y;
		q.pop();
		for (int i = head1[u]; i; i = next1[i]) {
			y = ver1[i];
			f[y].a = std::max(vv[y] + f[u].a, f[y].a);
			deg[y]--;
			if (!deg[y])
				q.push(y);
		}
	}
}

int cmp(node x, node y) {
	return x.a > y.a;
}

int main() {
	freopen("azeroth.in", "r", stdin);
	freopen("azeroth.out", "w", stdout);
	scanf("%d %d", &n, &m);
	for (int i = 1, a, b; i <= m; i++) {
		scanf("%d %d", &a, &b);
		if (a != b)
			add(a, b);
	}
	for (int i = 1; i <= n; i++)
		scanf("%d", &vw[i]);
	scanf("%d", &k);
	for (int i = 1; i <= n; i++)
		if (!dfn[i])
			tarjan(i);
	for (int i = 1; i <= n; i++)
		for (int j = head[i]; j; j = next[j])
			if (scc[i] != scc[ver[j]]) {
				int f1 = find(scc[i]), f2 = find(scc[ver[j]]);
				add1(scc[i], scc[ver[j]]);
				deg[scc[ver[j]]]++;
				fa[f1] = f2;
			}
	dp();
	std::sort(f + 1, f + col + 1, cmp);
	k++;
	for (int i = 1, f1; i <= col && k; i++)
		if (!used[f1 = find(f[i].id)]) {
			used[f1] = 1;
			ans += f[i].a;
			k--;
		}
	printf("%d", ans);
}

坑

并查集出锅
数组大小没开够（边与点）