裸的树链剖分加线段树区间修改
区间合并时需要多注意一点
当时写的很慢 理解不深刻

#include<bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 40005;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1 int N,M; struct Pode{
int to,next;
}edge[MAXN * 2];
int tot;
int head[MAXN];
int E[MAXN][3];
void addedge(int x,int y){
edge[tot].to = y; edge[tot].next = head[x]; head[x] = tot ++;
} int top[MAXN],fa[MAXN],son[MAXN],deep[MAXN],num[MAXN],p[MAXN],fp[MAXN],pos;
void dfs1(int x,int pre,int dep){
deep[x] = dep;
fa[x] = pre;
num[x] = 1;
for(int i = head[x]; i != -1; i = edge[i].next){
int y = edge[i].to;
if(y == pre) continue;
dfs1(y,x,dep + 1);
num[x] += num[y];
if(son[x] == -1 || num[y] > num[son[x]])
son[x] = y;
}
}
void dfs2(int x,int tp){
top[x] = tp;
p[x] = pos++;
fp[p[x]] = x;
if(son[x] == -1) return;
dfs2(son[x],tp);
for(int i = head[x] ; i != -1; i = edge[i].next){
int y = edge[i].to;
if(y != son[x] && y != fa[x])
dfs2(y,y);
}
}
int ininum[MAXN];
struct Node{
int l,r, num;
Node(int a=-1, int b=0, int c=0):l(a), r(b), num(c){}
Node operator + (const Node &T)const {
if(l == -1) return T; if(T.l == -1) return Node(l,r,num);
Node tt;
tt.l = l; tt.r = T.r;
tt.num = num+T.num;
if(r == T.l) tt.num--;
return tt;
}
Node rev(){
return Node(r,l,num);
}
};
struct Segtree{
Node tree[MAXN<<2];
int lazy[MAXN<<2];
void Pushup(int rt){
tree[rt] = tree[rt<<1]+tree[rt<<1|1];
}
void Pushdown(int rt) {
if(lazy[rt] != -1) {
lazy[rt<<1] = lazy[rt<<1|1] = lazy[rt];
tree[rt<<1|1] = tree[rt<<1] = Node(lazy[rt], lazy[rt], 1);
lazy[rt] = -1;
}
}
void Build(int l,int r,int rt) {
lazy[rt] = -1;
if(l == r) {
tree[rt] = Node(ininum[l], ininum[l], 1);
return;
}
int m = (l+r) >>1;
Build(lson); Build(rson);
Pushup(rt);
}
void Change(int L,int R,int num,int l,int r,int rt) {
if(L <= l && r <= R) {
tree[rt] = Node(num,num,1);
lazy[rt] = num;
return;
}
int m = (l+r) >>1;
Pushdown(rt);
if(L <= m) Change(L,R,num,lson);
if(R > m) Change(L,R,num,rson);
Pushup(rt);
} Node Sum(int L,int R,int l,int r,int rt){
if(L <= l && r <= R) {
return tree[rt];
}
int m = (l + r) >> 1;
Node ans;
Pushdown(rt);
if(L <= m) ans = ans+Sum(L,R,lson);
if(R > m) ans = ans+Sum(L,R,rson);
return ans;
}
void Find(int x,int y,int d){
int t1 = top[x]; int t2 = top[y];
while(t1 != t2) {
if(deep[t1] < deep[t2]) {
swap(t1,t2); swap(x,y);
}
Change(p[t1], p[x], d, 1,N,1);
x = fa[t1];
t1 = top[x];
}
if(x == y) return;
if(deep[x] > deep[y]) {
swap(x,y);
}
Change(p[son[x]], p[y], d,1, N, 1);
}
Node Query(int x,int y){
int t1 = top[x]; int t2 = top[y];
Node X, Y;
while(t1 != t2) {
if(deep[t1] < deep[t2]){
Y = Sum(p[t2], p[y], 1,N,1)+Y;
y = fa[t2]; t2 = top[y];
}else {
X = Sum(p[t1], p[x], 1,N,1)+X;
x = fa[t1]; t1 = top[x];
}
}
if(x == y) return X+Y;
if(deep[x] > deep[y]){
return Y.rev() + Sum(p[son[y]], p[x], 1,N,1) + X;
}else {
return X.rev() + Sum(p[son[x]], p[y], 1,N,1) + Y;
}
}
}solve; int main(){
while(~scanf("%d %d",&N,&M)){
memset(head,-1,sizeof(head));
memset(son, -1, sizeof(son));
tot = 0;
pos = 1;
for(int i = 1; i < N; ++i){
scanf("%d%d%d",&E[i][0],&E[i][1],&E[i][2]);
addedge(E[i][0], E[i][1]); addedge(E[i][1], E[i][0]);
}
dfs1(1,0,1);
dfs2(1,1); ininum[0] = 0;
for(int i = 1; i < N; ++i){
if(deep[E[i][0]] < deep[E[i][1]]) swap(E[i][0], E[i][1]);
ininum[ p[E[i][0]] ] = E[i][2];
}
solve.Build(1,N,1); for(int i = 1; i <= M; ++i) {
char a[10]; int b,c,d;
scanf("%s",a);
if(a[0] == 'C') {
scanf("%d %d %d",&b,&c,&d);
solve.Find(b,c,d);
}else {
scanf("%d %d",&b,&c);
printf("%d\n", solve.Query(b,c).num );
}
}
}
return 0;
}

hdu5893 List wants to travel的更多相关文章

  1. hdu5893 List wants to travel(树链剖分+线段树)

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submissi ...

  2. 图论 - Travel

    Travel The country frog lives in has nn towns which are conveniently numbered by 1,2,…,n. Among n(n− ...

  3. 【BZOJ-1576】安全路径Travel Dijkstra + 并查集

    1576: [Usaco2009 Jan]安全路经Travel Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 1044  Solved: 363[Sub ...

  4. Linux inode && Fast Directory Travel Method(undone)

    目录 . Linux inode简介 . Fast Directory Travel Method 1. Linux inode简介 0x1: 磁盘分割原理 字节 -> 扇区(sector)(每 ...

  5. HDU - Travel

    Problem Description Jack likes to travel around the world, but he doesn’t like to wait. Now, he is t ...

  6. 2015弱校联盟(1) - I. Travel

    I. Travel Time Limit: 3000ms Memory Limit: 65536KB The country frog lives in has n towns which are c ...

  7. ural 1286. Starship Travel

    1286. Starship Travel Time limit: 1.0 secondMemory limit: 64 MB It is well known that a starship equ ...

  8. Travel Problem[SZU_K28]

    DescriptionAfter SzuHope take part in the 36th ACMICPC Asia Chendu Reginal Contest. Then go to QingC ...

  9. hdu 5441 travel 离线+带权并查集

    Time Limit: 1500/1000 MS (Java/Others)  Memory Limit: 131072/131072 K (Java/Others) Problem Descript ...

随机推荐

  1. CF798E. Mike and code of a permutation [拓扑排序 线段树]

    CF798E. Mike and code of a permutation 题意: 排列p,编码了一个序列a.对于每个i,找到第一个\(p_j > p_i\)并且未被标记的j,标记这个j并\( ...

  2. [Sdoi2017]序列计数 [矩阵快速幂]

    [Sdoi2017]序列计数 题意:长为\(n \le 10^9\)由不超过\(m \le 2 \cdot 10^7\)的正整数构成的和为\(t\le 100\)的倍数且至少有一个质数的序列个数 总- ...

  3. CodeChef Little Elephant and Mouses [DP]

    https://www.codechef.com/problems/LEMOUSE 题意: 有一个n *m的网格.有一头大象,初始时在(1,1),要移动到(n,m),每次只能向右或者向下走.有些格子中 ...

  4. BZOJ 3907: 网格 [Catalan数 高精度]

    3907: 网格 Time Limit: 1 Sec  Memory Limit: 256 MBSubmit: 402  Solved: 180[Submit][Status][Discuss] De ...

  5. BZOJ 2083: [Poi2010]Intelligence test [vector+二分]

    2083: [Poi2010]Intelligence test Time Limit: 10 Sec  Memory Limit: 259 MBSubmit: 469  Solved: 227[Su ...

  6. Netty ByteBuf梳理

    我们知道,网络数据的基本单位总是字节.Java NIO提供了ByteBuffer作为它的字节容器,但是这个类使用起来过于复杂,而且也有些繁琐. Netty的ByteBuffer替代品是ByteBuf, ...

  7. HashMap中的散列函数、冲突解决机制和rehash

    一.概述 散列算法有两个主要的实现方式:开散列和闭散列,HashMap采用开散列实现. HashMap中,键值对(key-value)在内部是以Entry(HashMap中的静态内部类)实例的方式存储 ...

  8. mysql有多条记录的单个字段想存为一个字段显示的方法

    SELECT po.id,(SELECT GROUP_CONCAT(mr.member_type) as memberTypeList FROM prod_offer_member_rel mr WH ...

  9. Spring mvc学习指南

    使用flash attribute(闪存传值) 在配置文件中添加<mvc:annotion-driven/> 在controller方法参数里面添加RedirectAttributes r ...

  10. 关于c++栈溢出的问题

    我自己定义了一个数据类型node,嵌套在另一个数据类型当中时候,用到了delete函数, 在我node的声明当中声明了几个指针 在我的析构函数中却调用了delet函数 结果程序结果是能跑出来 提示我栈 ...