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;
}
原文地址:https://www.cnblogs.com/Tony-Double-Sky/p/9895580.html
时间: 2024-10-20 06:12:00