[ZJOI2008]骑士

看题时像一个独立集的题，看了数据之后。。。。

主要在于：每个骑士都有且仅有一个自己最厌恶的骑士。

所以整个图是多个联通块，每联通块都是一个环和一些树。

先在树上做dp，

可以枚举环上两点，分别不取。

也可以写个环dp。

（我写的是环dp）

/**
 * Problem:[ZJOI2008]p5
 * Author:Shun Yao
 * Time:2013.5.30
 * Result:Accepted
 * Memo:DP
 */

#include <cstring>
#include <cstdio>

const long Maxn = 1000005;

long long max(long long x, long long y) {
	return x > y ? x : y;
}

void swap(long &x, long &y) {
	x ^= y;
	y ^= x;
	x ^= y;
}

long n, A[Maxn], fa[Maxn], ff[Maxn];
long long ans, f[Maxn][2];

class gnode {
public:
	long v;
	gnode *next;
	gnode() {}
	~gnode() {}
	gnode(long V, gnode *ne):v(V), next(ne) {}
} *g[Maxn];

void add(long x, long y) {
	g[x] = new gnode(y, g[x]);
	g[y] = new gnode(x, g[y]);
}

void DP_Tree(long x) {
	static long q[Maxn], *l, *r;
	static gnode *e;
	r = (l = &(*q = x)) + 1;
	while (l != r) {
		for (e = g[*l]; e; e = e->next)	if (!ff[e->v]) {
			ff[e->v] = *l;
			*(r++) = e->v;
		}
		++l;
	}
	do {
		--l;
		f[*l][0] = 0;
		f[*l][1] = A[*l];
		for (e = g[*l]; e; e = e->next) if (ff[e->v] == *l) {
			f[*l][0] += f[e->v][1];
			f[*l][1] += f[e->v][0];
		}
		f[*l][1] = max(f[*l][1], f[*l][0]);
	} while (l != q);
}

long q[Maxn], *l, *r;
long st[Maxn << 1], stl;
long ST[Maxn], STl;
long d[Maxn];
long long h[Maxn];
void DP(long x) {
	static gnode *e;
	static long a, b;
	r = (l = &(*q = x)) + 1;
	fa[x] = -1;
	stl = 0;
	STl = 0;
	while (l != r) {
		for (e = g[*l]; e; e = e->next) if (e->v != fa[*l]) {
			if (!fa[e->v]) {
				d[e->v] = d[*l] + 1;
				fa[e->v] = *l;
				*(r++) = e->v;
			} else if (!stl) {
				a = *l;
				b = e->v;
				if (d[a] < d[b])
					swap(a, b);
				while (d[a] > d[b]) {
					st[++stl] = a;
					a = fa[a];
				}
				while (a != b) {
					st[++stl] = a;
					ST[++STl] = b;
					a = fa[a];
					b = fa[b];
				}
				st[++stl] = a;
				while (STl)
					st[++stl] = ST[STl--];
			}
		}
		++l;
	}
	static long i;
	for (i = 1; i <= stl; ++i)
		ff[st[i]] = -1;
	for (i = 1; i <= stl; ++i)
		DP_Tree(st[i]);
	static long long ret;
	h[0] = 0;
	h[1] = f[st[1]][1];
	for (i = 2; i < stl; ++i)
		h[i] = max(h[i - 2] + f[st[i - 1]][0] + f[st[i]][1], h[i - 1] + f[st[i]][0]);
	ret = h[stl - 1] + f[st[stl]][0];
	h[1] = 0;
	h[2] = f[st[2]][1];
	for (i = 3; i <= stl; ++i)
		h[i] = max(h[i - 2] + f[st[i - 1]][0] + f[st[i]][1], h[i - 1] + f[st[i]][0]);
	ans += max(h[stl] + f[st[1]][0], ret);
}

int main() {
	static long i, B;
	freopen("p5.in", "r", stdin);
	freopen("p5.out", "w", stdout);
	scanf("%ld", &n);
	for (i = 1; i <= n; ++i)
		g[i] = 0;
	for (i = 1; i <= n; ++i) {
		scanf("%ld%ld", A + i, &B);
		add(i, B);
	}
	memset(d, 0, sizeof d);
	memset(fa, 0, sizeof fa);
	memset(ff, 0, sizeof ff);
	ans = 0;
	for (i = 1; i <= n; ++i)
		if (!fa[i])
			DP(i);
	printf("%lld", ans);
	fclose(stdin);
	fclose(stdout);
	return 0;
}