[树链剖分]

[BZOJ 3531][SDOI 2014]旅行

动态开点线段树+树剖。

写完 编译 一遍过样例 交了 1A 这感觉还挺爽QAQ。。

#include <bits/stdc++.h>
#define maxn 100010
using namespace std;

int n, Q;
int w[maxn], c[maxn];
//-------------------------------------------------------------------//
struct Edge{
	int to, next;
}edge[maxn << 1];
int h[maxn], cnt;
void add(int u, int v){
	cnt ++;
	edge[cnt].to = v;
	edge[cnt].next = h[u];
	h[u] = cnt;
}
//-------------------------------------------------------------------//
int son[maxn], top[maxn], pos[maxn], dfs_clock, size[maxn];
int fa[maxn], dep[maxn];

void dfs1(int u){
	dep[u] = dep[fa[u]] + 1;
	size[u] = 1;
	for(int i = h[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if(v == fa[u])continue;
		fa[v] = u;
		dfs1(v);
		size[u] += size[v];
		if(size[v] > size[son[u]])
		    son[u] = v;
	}
}

void dfs2(int u, int tp){
	pos[u] = ++ dfs_clock;
	top[u] = tp;
	if(son[u])dfs2(son[u], tp);
	for(int i = h[u]; i; i = edge[i].next){
		int v = edge[i].to;
		if(v == fa[u] || v == son[u])
		    continue;
		dfs2(v, v);
	}
}

//-------------------------------------------------------------------//

#define M 10000010
int root[maxn], lc[M], rc[M], mx[M], s[M], Smz;

void pushup(int id){
	s[id] = s[lc[id]] + s[rc[id]];
	mx[id] = max(mx[lc[id]], mx[rc[id]]);
}

void update(int &id, int l, int r, int p, int val){
	if(id == 0)id = ++ Smz;
	if(l == r){s[id] = val, mx[id] = val;return;}
	int mid = l + r >> 1;
	if(p <= mid)update(lc[id], l, mid, p, val);
	else update(rc[id], mid+1, r, p, val);
	pushup(id);
}

int Asksum(int id, int l, int r, int L, int R){
	if(id == 0)return 0;
	if(l == L && R == r)return s[id];
	int mid = l + r >> 1;
	if(R <= mid)return Asksum(lc[id], l, mid, L, R);
	if(L > mid) return Asksum(rc[id], mid+1, r, L, R);
	return Asksum(lc[id], l, mid, L, mid) + Asksum(rc[id], mid+1, r, mid+1, R);
}

int Askmx(int id, int l, int r, int L, int R){
	if(id == 0)return 0;
	if(l == L && R == r)return mx[id];
	int mid = l + r >> 1;
	if(R <= mid)return Askmx(lc[id], l, mid, L, R);
	if(L > mid) return Askmx(rc[id], mid+1, r, L, R);
	return max(Askmx(lc[id], l, mid, L, mid), Askmx(rc[id], mid+1, r, mid+1, R));
}

int asksum(int u, int v){
	int C = c[u], ret = 0;
	while(top[u] != top[v]){
		if(dep[top[u]] < dep[top[v]])
		    swap(u, v);
		ret += Asksum(root[C], 1, n, pos[top[u]], pos[u]);
		u = fa[top[u]];
	}
	if(dep[u] > dep[v])swap(u, v);
	ret += Asksum(root[C], 1, n, pos[u], pos[v]);
	return ret;
}

int askmx(int u, int v){
    int C = c[u], ret = 0;
	while(top[u] != top[v]){
		if(dep[top[u]] < dep[top[v]])
		    swap(u, v);
		ret = max(ret, Askmx(root[C], 1, n, pos[top[u]], pos[u]));
		u = fa[top[u]];
	}
	if(dep[u] > dep[v])swap(u, v);
	ret = max(ret, Askmx(root[C], 1, n, pos[u], pos[v]));
	return ret;
}

int main(){
	char cmd[2];
	scanf("%d%d", &n, &Q);
	for(int i = 1; i <= n; i ++)
	    scanf("%d%d", &w[i], &c[i]);
	int x, y;
	for(int i = 1; i < n; i ++){
		scanf("%d%d", &x, &y);
		add(x, y), add(y, x);
	}
	
	dfs1(1);
	dfs2(1, 1);
	for(int i = 1; i <= n; i ++)
		update(root[c[i]], 1, n, pos[i], w[i]);
	
	while(Q --){
		scanf("%s", cmd);
		scanf("%d%d", &x, &y);
		switch(cmd[1]){
			case 'C':
                update(root[c[x]], 1, n, pos[x], 0), c[x] = y;
                update(root[y], 1, n, pos[x], w[x]);
				break;
			case 'W':
				update(root[c[x]], 1, n, pos[x], w[x] = y);
				break;
			case 'S':
				printf("%d\n", asksum(x, y));
				break;
			case 'M':
                printf("%d\n", askmx(x, y));
				break;
		}
	}
	return 0;
}