[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;

}