HDU 5044 Tree 树链剖分+区间标记

Tree

Problem Description

You are given a tree (an acyclic undirected connected graph) with N nodes. The tree nodes are numbered from 1 to N

There are N - 1 edges numbered from 1 to N - 1.

Each node has a value and each edge has a value. The initial value is 0.

There are two kind of operation as follows:

● ADD1 u v k: for nodes on the path from u to v, the value of these nodes increase by k.

● ADD2 u v k: for edges on the path from u to v, the value of these edges increase by k.

After finished M operation on the tree, please output the value of each node and edge.

 

Input

The first line of the input is T (1 ≤ T ≤ 20), which stands for the number of test cases you need to solve.

The first line of each case contains two integers N ,M (1 ≤ N, M ≤10⁵),denoting the number of nodes and operations, respectively.

The next N - 1 lines, each lines contains two integers u, v(1 ≤ u, v ≤ N ), denote there is an edge between u,v and its initial value is 0.

For the next M line, contain instructions “ADD1 u v k” or “ADD2 u v k”. (1 ≤ u, v ≤ N, -10⁵ ≤ k ≤ 10⁵)

 

Output

For each test case, print a line “Case #t:”(without quotes, t means the index of the test case) at the beginning.

The second line contains N integer which means the value of each node.

The third line contains N - 1 integer which means the value of each edge according to the input order.

 

Sample Input

2 4 2 1 2 2 3 2 4 ADD1 1 4 1 ADD2 3 4 2 4 2 1 2 2 3 1 4 ADD1 1 4 5 ADD2 3 2 4

 

Sample Output

Case #1: 1 1 0 1 0 2 2 Case #2: 5 0 0 5 0 4 0

 

题意：

　　给你一棵树，n个结点

　　1、将u到v之间的结点权值加k；

　　2、将u到v之间的边权加k。

　　输出经过修改后所有的边权和点权。

题解：

　　这题线段树超时

　　需要用到一个区间更新小技巧

　　比如x,y这个区间+k，那么sum[x] += k, sum[y+1] -= k;

　　最后统计一个前缀和就可以了

　　这题读入挂超时，不用能A，格式也要注意了

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 2e5+10, M = 1e3+20, mod = 1e9+7, inf = 2e9;

int dep[N],head[N],t=1,sz[N],fa[N],indexS,top[N],pos[N],son[N],idpos[N];
struct ss{int to,next,id;}e[N*2];
int n;
void add(int u,int v,int id)
{e[t].to = v;e[t].next = head[u];e[t].id=id;head[u] = t++;}
void dfs(int u) {
    int k = 0;
    sz[u] = 1;
    dep[u] = dep[fa[u]] + 1;
    for(int i = head[u]; i; i = e[i].next) {
        int to = e[i].to;
        if(to == fa[u]) continue;
        fa[to] = u;
        idpos[to] = e[i].id;
        dfs(to);
        sz[u] += sz[to];
        if(sz[to] > sz[k]) k = to;
    }
    if(k) son[u] = k;
}
void dfs(int u,int chain) {
    int k = 0;
    pos[u] = ++indexS;
    top[u] = chain;
    if(son[u] > 0)
        dfs(son[u],chain);
    for(int i = head[u]; i; i = e[i].next) {
        int to = e[i].to;
        if(dep[to] > dep[u] && son[u] != to)
            dfs(to,to);
    }
}
LL sum[N],sum2[N];
void updateL(int x,int y,int k) {
    while(top[x] != top[y]) {
        if(dep[top[x]] < dep[top[y]]) swap(x,y);
        sum2[pos[top[x]]] +=k;
        sum2[pos[x]+1] -=k;
        x = fa[top[x]];
    }
    if(dep[x] < dep[y]) swap(x,y);
    sum2[pos[y]] +=k;
    sum2[pos[x]+1] -=k;
}
void updateW(int x,int y,int k) {
    while(top[x] != top[y]) {
        if(dep[top[x]] < dep[top[y]]) swap(x,y);
        sum[pos[top[x]]] += k;
        sum[pos[x] + 1] -= k;
        x = fa[top[x]];
    }
    if(dep[x] > dep[y]) swap(x,y);
    sum[pos[x] + 1] += k;
    sum[pos[y] + 1] -= k;
}
void init() {
    t = 1;
    memset(head,0,sizeof(head));
    indexS = 0;fa[1] = 0;
    for(int i = 0; i <= n+5; ++i) sum[i] = 0, sum2[i] = 0;
    for(int i = 0; i <= n; ++i) son[i] = 0,sz[i] = 0;
}
int T,m,cas = 1;
LL f[N];
int main() {
    scanf("%d",&T);
    while(T--) {
      scanf("%d%d",&n,&m);
        init();
        for(int i = 1 ; i < n; ++i) {
            int x,y;
            scanf("%d%d",&x,&y);
            add(x,y,i),add(y,x,i);
        }
        dfs(1);
        dfs(1,1);
        while(m--) {
            char ch[10];
            int x,y,k;
            scanf("%s",ch);
           scanf("%d%d%d",&x,&y,&k);
            if(ch[3] == '1') {
                updateL(x,y,k);
            }
            else if(x!=y){
                updateW(x,y,k);
            }
        }
        printf("Case #%d:
",cas++);
        for(int i = 2; i <= n; ++i)
            sum[i]+=sum[i-1],sum2[i]+=sum2[i-1];
        for(int i = 2; i <= n; ++i) f[idpos[i]] = sum[pos[i]];
        for(int i = 1; i < n; ++i)
            printf("%I64d ",sum2[pos[i]]);
        printf("%I64d
",sum2[pos[n]]);
        for(int i = 1; i < n-1; ++i) printf("%I64d ",f[i]);
        if(n-1 > 0)printf("%I64d",f[n-1]);
        puts("");
    }
    return 0;
}