hdu5029 Relief grain (树链剖分)

Relief grain

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 100000/100000 K (Java/Others)
Total Submission(s): 686 Accepted Submission(s): 163

Problem Description

The soil is cracking up because of the drought and the rabbit kingdom is facing a serious famine. The RRC(Rabbit Red Cross) organizes the distribution of relief grain in the disaster area.

We can regard the kingdom as a tree with n nodes and each node stands for a village. The distribution of the relief grain is divided into m phases. For each phases, the RRC will choose a path of the tree and distribute some relief grain of a certain type for every village located in the path.

There are many types of grains. The RRC wants to figure out which type of grain is distributed the most times in every village.

Input

The input consists of at most 25 test cases.

For each test case, the first line contains two integer n and m indicating the number of villages and the number of phases.

The following n-1 lines describe the tree. Each of the lines contains two integer x and y indicating that there is an edge between the x-th village and the y-th village.

The following m lines describe the phases. Each line contains three integer x, y and z indicating that there is a distribution in the path from x-th village to y-th village with grain of type z. (1 <= n <= 100000, 0 <= m <= 100000, 1 <= x <= n, 1 <= y <= n, 1 <= z <= 100000)

The input ends by n = 0 and m = 0.

Output

For each test case, output n integers. The i-th integer denotes the type that is distributed the most times in the i-th village. If there are multiple types which have the same times of distribution, output the minimal one. If there is no relief grain in a village, just output 0.

Sample Input

2 4 1 2 1 1 1 1 2 2 2 2 2 2 2 1 5 3 1 2 3 1 3 4 5 3 2 3 3 1 5 2 3 3 3 0 0

Sample Output

1 2 2 3 3 0 2

Hint

For the first test case, the relief grain in the 1st village is {1, 2}, and the relief grain in the 2nd village is {1, 2, 2}.

Source

2014 ACM/ICPC Asia Regional Guangzhou Online

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解题思路：这道题想通的话很简单，对数值建线段树，由于是单点查询，对于一个区间[l,r]置z来说，它的影响就是在l

位置把z的次数+1，然后在r+1的位置把z的次数-1，所以通过树链剖分把区间拆成logn段，再把每段的左右点的z记录下来，最后从左往右扫一遍，把这个点所有z的记录都单点更新线段树就行了，因为只有mlogn段，所以总的插入的效率不会超过2m*logn*logn。

比赛时没做出来真是too young too naive。。。

#include 
#include 
#include 
#define maxn 100010
using namespace std;
int n,m;
vector G[maxn],pl[maxn],pr[maxn];
int top[maxn],w[maxn],tw[maxn],pre[maxn],tn,son[maxn],dep[maxn],num[maxn],ans[maxn];
void init(){
    for(int i=0;inum[son[u]]) son[u]=v;
        }
    }
}
void build_tree(int u,int fa){
    w[u]=++tn,tw[tn]=u,top[u]=fa;
    if(son[u]) build_tree(son[u],fa);
    int nn=G[u].size();
    for(int i=0;i=a[k<<1|1].Max){
        a[k].v=a[k<<1].v;
        a[k].Max=a[k<<1].Max;
    }else{
        a[k].v=a[k<<1|1].v;
        a[k].Max=a[k<<1|1].Max;
    }
}
void build(int l,int r,int k){
    a[k].l=l,a[k].r=r;
    if(l==r){
        a[k].v=l;
        a[k].Max=0;
    }else{
        int mid=(l+r)>>1;
        build(l,mid,k<<1);
        build(mid+1,r,k<<1|1);
        pushup(k);
    }
}

void insert(int x,int v,int k){
    if(a[k].l==a[k].r){
        a[k].Max+=v;
    }else{
        int mid=(a[k].l+a[k].r)>>1;
        if(x<=mid) insert(x,v,k<<1);
        else insert(x,v,k<<1|1);
        pushup(k);
    }
}
void read(){
    int u,v;
    for(int i=0;i