P4315 月下“毛景树”

P4315 月下“毛景树”

题目描述
毛毛虫经过及时的变形,最终逃过的一劫,离开了菜妈的菜园。 毛毛虫经过千山万水,历尽千辛万苦,最后来到了小小的绍兴一中的校园里。

爬啊爬爬啊爬毛毛虫爬到了一颗小小的“毛景树”下面,发现树上长着他最爱吃的毛毛果 “毛景树”上有N个节点和N-1条树枝,但节点上是没有毛毛果的,毛毛果都是长在树枝上的。但是这棵“毛景树”有着神奇的魔力,他能改变树枝上毛毛果的个数:

Change k w:将第k条树枝上毛毛果的个数改变为w个。

Cover u v w:将节点u与节点v之间的树枝上毛毛果的个数都改变为w个。

Add u v w:将节点u与节点v之间的树枝上毛毛果的个数都增加w个。 由于毛毛虫很贪,于是他会有如下询问:

Max u v:询问节点u与节点v之间树枝上毛毛果个数最多有多少个。

Solution

树剖
注意线段树下推标记时分清楚对子节点标记的影响

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#define LL long long
#define REP(i, x, y) for(int i = (x);i <= (y);i++)
using namespace std;
int RD(){
    int out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const int maxn = 100019,INF = 1e9 + 19;
int head[maxn],nume = 1;
struct Node{
    int v,dis,nxt;
    }E[maxn << 3];
void add(int u,int v,int dis){
    E[++nume].nxt = head[u];
    E[nume].v = v;
    E[nume].dis = dis;
    head[u] = nume;
    }
int num;
int size[maxn], wson[maxn], dep[maxn], fa[maxn], val[maxn];
int top[maxn], pos[maxn], ori[maxn], tot;
void dfs1(int u, int F){
	size[u] = 1;
	for(int i = head[u];i;i = E[i].nxt){
		int v = E[i].v;
		if(v == F)continue;
		val[v] = E[i].dis;
		dep[v] = dep[u] + 1;
		fa[v] = u;
		dfs1(v, u);
		size[u] += size[v];
		if(size[v] > size[wson[u]])wson[u] = v;
		}
	}
void dfs2(int u, int TP){
	top[u] = TP;
	pos[u] = ++tot;
	ori[tot] = u;
	if(!wson[u])return ;
	dfs2(wson[u], TP);
	for(int i = head[u];i;i = E[i].nxt){
		int v = E[i].v;
		if(v == fa[u] || v == wson[u])continue;
		dfs2(v, v);
		}
	}
#define lid (id << 1)
#define rid (id << 1) | 1
struct seg_tree{
	int l, r;
	int max;
	int add, set;
	}tree[maxn << 2];
void pushup(int id){tree[id].max = max(tree[lid].max, tree[rid].max);}
void build(int id, int l, int r){
	tree[id].l = l, tree[id].r = r, tree[id].set = -1;
	if(l == r){
		tree[id].max = val[ori[l]];
		return ;
		}
	int mid = (l + r) >> 1;
	build(lid, l, mid), build(rid, mid + 1, r);
	pushup(id);
	}
void pushdown(int id){
	if(tree[id].set != -1){
		int v = tree[id].set;
		tree[lid].max = tree[rid].max = v;
		tree[lid].set = tree[rid].set = v;
		tree[lid].add = tree[rid].add = 0;
		tree[id].add = 0;
		tree[id].set = -1;
		}
	if(tree[id].add != 0){
		int v = tree[id].add;
		tree[lid].max += v;
		tree[rid].max += v;
		if(tree[lid].set != -1)tree[lid].set += v;
		if(tree[rid].set != -1)tree[rid].set += v;
		tree[lid].add += v;
		tree[rid].add += v;
		tree[id].add = 0;
		}
	}
void update(int id, int v, int l, int r, int o){
	pushdown(id);
	if(tree[id].l == l && tree[id].r == r){
		if(o == 1){//1为全改变
			tree[id].max = v;
			tree[id].set = v;
			}
		else{//0为区间加
			tree[id].max += v;
			tree[id].add += v;
			}
		return ;
		}
	int mid = (tree[id].l + tree[id].r) >> 1;
	if(mid < l)update(rid, v, l, r, o);
	else if(mid >= r)update(lid, v, l, r, o);
	else update(lid, v, l, mid, o), update(rid, v, mid + 1, r, o);
	pushup(id);
	}
int query(int id, int l, int r){
	pushdown(id);
	if(tree[id].l == l && tree[id].r == r)return tree[id].max;
	int mid = (tree[id].l + tree[id].r) >> 1;
	if(mid < l)return query(rid, l, r);
	else if(mid >= r)return query(lid, l, r);
	else return max(query(lid, l, mid), query(rid, mid + 1, r));
	}
void uprange(int x, int y, int v, int o){
	while(top[x] != top[y]){
		if(dep[top[x]] < dep[top[y]])swap(x, y);
		update(1, v, pos[top[x]], pos[x], o);
		x = fa[top[x]];
		}
	if(x == y)return ;
	if(dep[x] > dep[y])swap(x, y);
	update(1, v, pos[x] + 1, pos[y], o);
	}
int Q_max(int x, int y){
	int ret = 0;
	while(top[x] != top[y]){
		if(dep[top[x]] < dep[top[y]])swap(x, y);
		ret = max(ret, query(1, pos[top[x]], pos[x]));
		x = fa[top[x]];
		}
	if(x == y)return ret;
	if(dep[x] > dep[y])swap(x, y);
	ret = max(ret, query(1, pos[x] + 1, pos[y]));
	return ret;
	}
struct EDG{int x, y;}I[maxn];
void init(){
	num = RD();
	REP(i, 1, num - 1){
		int u = RD(), v = RD(), dis = RD();
		I[i] = (EDG){u, v};
		add(u, v, dis), add(v, u, dis);
		}
	dep[1] = 1;
	dfs1(1, -1);
	dfs2(1, 1);
	build(1, 1, num);
	}
void solve(){
	char cmd[19];
	while(1){
		scanf("%s", cmd);
		if(cmd[0] == 'S')return ;
		else if(cmd[0] == 'C' && cmd[1] == 'h'){
			int k = RD(), w = RD();
			uprange(I[k].x, I[k].y, w, 1);
			}
		else if(cmd[0] == 'C' && cmd[1] == 'o'){
			int x = RD(), y = RD(), w = RD();
			uprange(x, y, w, 1);
			}
		else if(cmd[0] == 'A'){
			int x = RD(), y = RD(), w = RD();
			uprange(x, y, w, 0);
			}
		else{
			int x = RD(), y = RD();
			printf("%d\n", Q_max(x, y));
			}
		}
	}
int main(){
	init();
	solve();
	return 0;
	}
posted @ 2018-11-02 12:24  Tony_Double_Sky  阅读(186)  评论(0编辑  收藏  举报