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POJ 3237 Tree

题目链接

就多一个取负操作，所以线段树结点就把最大和最小值存下来，每次取负的时候，最大和最小值取负后，交换即可

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 10005;
const int INF = 0x3f3f3f3f;
int dep[N], fa[N], son[N], sz[N], top[N], id[N], idx, val[N];
int first[N], next[N * 2], vv[N * 2], en;

struct Edge {
	int u, v, val;
	Edge() {}
	Edge(int u, int v, int val) {
		this->u = u;
		this->v = v;
		this->val = val;
	}
} e[N];

void init() {
	en = 0;
	idx = 0;
	memset(first, -1, sizeof(first));
}

void add_Edge(int u, int v) {
	vv[en] = v;
	next[en] = first[u];
	first[u] = en++;
}

void dfs1(int u, int f, int d) {
	dep[u] = d;
	sz[u] = 1;
	fa[u] = f;
	son[u] = 0;
	for (int i = first[u]; i + 1; i = next[i]) {
		int v = vv[i];
		if (v == f) continue;
		dfs1(v, u, d + 1);
		sz[u] += sz[v];
		if (sz[son[u]] < sz[v])
			son[u] = v;
	}
}

void dfs2(int u, int tp) {
	id[u] = ++idx;
	top[u] = tp;
	if (son[u]) dfs2(son[u], tp);
	for (int i = first[u]; i + 1; i = next[i]) {
		int v = vv[i];
		if (v == fa[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}

#define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2)

struct Node {
	int l, r, Max, Min;
	bool neg;
} node[N * 4];

void pushup(int x) {
	node[x].Max = max(node[lson(x)].Max, node[rson(x)].Max);
	node[x].Min = min(node[lson(x)].Min, node[rson(x)].Min);
}

void pushdown(int x) {
	if (node[x].neg) {
		node[lson(x)].Max = -node[lson(x)].Max;
		node[rson(x)].Max = -node[rson(x)].Max;
		node[lson(x)].Min = -node[lson(x)].Min;
		node[rson(x)].Min = -node[rson(x)].Min;
		swap(node[lson(x)].Max, node[lson(x)].Min);
		swap(node[rson(x)].Max, node[rson(x)].Min);
		node[lson(x)].neg = !node[lson(x)].neg;
		node[rson(x)].neg = !node[rson(x)].neg;
		node[x].neg = false;
	}
}

void build(int l, int r, int x = 0) {
	node[x].l = l; node[x].r = r; node[x].neg = false;
	if (l == r) {
		node[x].Max = node[x].Min = val[l];
		return;
	}
	int mid = (l + r) / 2;
	build(l, mid, lson(x));
	build(mid + 1, r, rson(x));
	pushup(x);
}

void change(int v, int val, int x = 0) {
	if (node[x].l == node[x].r) {
		node[x].Max = node[x].Min = val;
		return;
	}
	int mid = (node[x].l + node[x].r) / 2;
	pushdown(x);
	if (v <= mid) change(v, val, lson(x));
	if (v > mid) change(v, val, rson(x));
	pushup(x);
}

void negate(int l, int r, int x = 0) {
	if (node[x].l >= l && node[x].r <= r) {
		node[x].neg = !node[x].neg;
		node[x].Max = -node[x].Max;
		node[x].Min = -node[x].Min;
		swap(node[x].Max, node[x].Min);
		return;
	}
	int mid = (node[x].l + node[x].r) / 2;
	pushdown(x);
	if (l <= mid) negate(l, r, lson(x));
	if (r > mid) negate(l, r, rson(x));
	pushup(x);
}

int query(int l, int r, int x = 0) {
	if (node[x].l >= l && node[x].r <= r) {
		return node[x].Max;
	}
	int mid = (node[x].l + node[x].r) / 2;
	pushdown(x);
	int ans = -INF;
	if (l <= mid) ans = max(ans, query(l, r, lson(x)));
	if (r > mid) ans = max(ans, query(l, r, rson(x)));
	pushup(x);
	return ans;
}

void gao1(int u, int v) {
	int tp1 = top[u], tp2 = top[v];
	while (tp1 != tp2) {
		if (dep[tp1] < dep[tp2]) {
			swap(tp1, tp2);
			swap(u, v);
		}
		negate(id[tp1], id[u]);
		u = fa[tp1];
		tp1 = top[u];
	}
	if (u == v) return;
	if (dep[u] > dep[v]) swap(u, v);
	negate(id[son[u]], id[v]);
}

int gao2(int u, int v) {
	int ans = -INF;
	int tp1 = top[u], tp2 = top[v];
	while (tp1 != tp2) {
		if (dep[tp1] < dep[tp2]) {
			swap(tp1, tp2);
			swap(u, v);
		}
		ans = max(ans, query(id[tp1], id[u]));
		u = fa[tp1];
		tp1 = top[u];
	}
	if (u == v) return ans;
	if (dep[u] > dep[v]) swap(u, v);
	ans = max(ans, query(id[son[u]], id[v]));
	return ans;
}

int T, n;

int main() {
	scanf("%d", &T);
	while (T--) {
		init();
		scanf("%d", &n);
		int u, v, w;
		for (int i = 1; i < n; i++) {
			scanf("%d%d%d", &u, &v, &w);
			add_Edge(u, v);
			add_Edge(v, u);
			e[i] = Edge(u, v, w);
		}
		dfs1(1, 0, 1);
		dfs2(1, 1);
		for (int i = 1; i < n; i++) {
			if (dep[e[i].u] < dep[e[i].v]) swap(e[i].u, e[i].v);
			val[id[e[i].u]] = e[i].val;
		}
		build(1, idx);
		char Q[10];
		int a, b;
		while (scanf("%s", Q)) {
			if (Q[0] == 'D') break;
			scanf("%d%d", &a, &b);
			if (Q[0] == 'Q') printf("%d\n", gao2(a, b));
			if (Q[0] == 'C') change(id[e[a].u], b);
			if (Q[0] == 'N') gao1(a, b);
		}
	}
	return 0;
}

POJ 3723 Tree（树链剖分）

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原文地址：http://blog.csdn.net/accelerator_/article/details/39717029