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HYSBZ 2243 染色

题目链接

树链剖分，关键在于线段树的维护，对于每个结点要记录下最左边和最右边的颜色，合并的时候，如果颜色相同那么颜色段要减1

代码：

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

const int N = 100005;

int dep[N], fa[N], son[N], sz[N], top[N], id[N], idx, val[N], val2[N];
int first[N], next[N * 2], vv[N * 2], en;

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, lc, rc, cnt, setv;
	Node() {cnt = 0;}
} node[N * 4];

void pushup(int x) {
	node[x].cnt = node[lson(x)].cnt + node[rson(x)].cnt;
	if (node[lson(x)].rc == node[rson(x)].lc) node[x].cnt--;
	node[x].lc = node[lson(x)].lc;
	node[x].rc = node[rson(x)].rc;
}

void pushdown(int x) {
	if (node[x].setv == -1) return;
	node[lson(x)].setv = node[lson(x)].lc = node[lson(x)].rc = node[x].setv;
	node[rson(x)].setv = node[rson(x)].lc = node[rson(x)].rc = node[x].setv;
	node[lson(x)].cnt = node[rson(x)].cnt = 1;
	node[x].setv = -1;
}

void build(int l, int r, int x = 0) {
	node[x].l = l; node[x].r = r;
	node[x].setv = -1;
	if (l == r) {
		node[x].lc = node[x].rc = val[l];
		node[x].cnt = 1;
		return;
	}
	int mid = (l + r) / 2;
	build(l, mid, lson(x));
	build(mid + 1, r, rson(x));
	pushup(x);
}

void add(int l, int r, int v, int x = 0) {
	if (node[x].l >= l && node[x].r <= r) {
		node[x].cnt = 1;
		node[x].setv = node[x].lc = node[x].rc = v;
		return;
	}
	pushdown(x);
	int mid = (node[x].l + node[x].r) / 2;
	if (l <= mid) add(l, r, v, lson(x));
	if (r > mid) add(l, r, v, rson(x));
	pushup(x);
}

Node merge(Node a, Node b) {
	Node ans;
	if (a.cnt == 0) return b;
	if (b.cnt == 0) return a;
	ans.lc = a.lc;
	ans.rc = b.rc;
	ans.cnt = a.cnt + b.cnt;
	if (a.rc == b.lc) ans.cnt--;
	return ans;
}

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

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

int gao2(int u, int v) {
	int tp1 = top[u], tp2 = top[v];
	Node ans1, ans2;
	while (tp1 != tp2) {
		if (dep[tp1] < dep[tp2]) {
			swap(tp1, tp2);
			swap(u, v);
			swap(ans1, ans2);
		}
		ans1 = merge(query(id[tp1], id[u]), ans1);
		u = fa[tp1];
		tp1 = top[u];
	}
	if (dep[u] > dep[v]) {
		swap(ans1, ans2);
		swap(u, v);
	}
	ans2 = merge(query(id[u], id[v]), ans2);
	if (ans1.cnt == 0) return ans2.cnt;
	if (ans2.cnt == 0) return ans1.cnt;
	int ans = ans1.cnt + ans2.cnt;
	if (ans1.lc == ans2.lc) ans--;
	return ans;
}

int n, m;

int main() {
	while (~scanf("%d%d", &n, &m)) {
		init();
		for (int i = 1; i <= n; i++) scanf("%d", &val2[i]);
		int u, v;
		for (int i = 1; i < n; i++) {
			scanf("%d%d", &u, &v);
			add_Edge(u, v);
			add_Edge(v, u);
		}
		dfs1(1, 0, 1);
		dfs2(1, 1);
		for (int i = 1; i <= n; i++)
			val[id[i]] = val2[i];
		build(1, idx);
		char Q[10];
		int a, b, c;
		while (m--) {
			scanf("%s%d%d", Q, &a, &b);
			if (Q[0] == 'Q') printf("%d\n", gao2(a, b));
			else {
				scanf("%d", &c);
				gao1(a, b, c);
			}
		}
	}
	return 0;
}

HYSBZ 2243 染色 （树链剖分）

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原文地址：http://blog.csdn.net/accelerator_/article/details/39737769