2014多校第四场1006 HDU 4902 Nice boat 线段树 区间更新问题

http://acm.hdu.edu.cn/showproblem.php?pid=4902

Problem Description

There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli.

Let us continue our story, z*p(actually you) defeat the 'MengMengDa' party's leader, and the 'MengMengDa' party dissolved. z*p becomes the most famous guy among the princess's knight party. 

One day, the people in the party find that z*p has died. As what he has done in the past, people just say 'Oh, what a nice boat' and don't care about why he died.

Since then, many people died but no one knows why and everyone is fine about that. Meanwhile, the devil sends her knight to challenge you with Algorithm contest.

There is a hard data structure problem in the contest:

There are n numbers a_1,a_2,...,a_n on a line, everytime you can change every number in a segment [l,r] into a number x(type 1), or change every number a_i in a segment [l,r] which is bigger than x to gcd(a_i,x) (type 2).

You should output the final sequence.

 

Input

The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains a integers n.
The next line contains n integers a_1,a_2,...,a_n separated by a single space.
The next line contains an integer Q, denoting the number of the operations.
The next Q line contains 4 integers t,l,r,x. t denotes the operation type.

T<=2,n,Q<=100000
a_i,x >=0
a_i,x is in the range of int32(C++)

 

Output

For each test case, output a line with n integers separated by a single space representing the final sequence.
Please output a single more space after end of the sequence

 

Sample Input

   
   
   
   

    
    
    
    1
10
16807 282475249 1622650073 984943658 1144108930 470211272 101027544 1457850878 1458777923 2007237709 
10
1 3 6 74243042
2 4 8 16531729
1 3 4 1474833169
2 1 8 1131570933
2 7 9 1505795335
2 3 7 101929267
1 4 10 1624379149
2 2 8 2110010672
2 6 7 156091745
1 2 5 937186357
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    16807 937186357 937186357 937186357 937186357 1 1 1624379149 1624379149 1624379149 
   
   
   
   

线段树的题目，区间更新，延迟标记。

连着两场多校比赛出现了线段树的题目，这是下半年亚洲赛要出题的节奏吗？哈哈~~~

解题思路：够造线段树，每个顶点包含两个值，一个flag作为标记的标志，另一个value做为节点的存储值，如果该节点左右子节点的值是相同的那么我们就把该节点的值也赋为它，若不相同就把其值赋为-1.标记2时只标记到子节点相同的点为止。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=110005;
int a[maxn];
struct SegementTree
{
    struct Tree
    {
        int l,r;
        int flag;
        int value;
    } tree[maxn*4];
    int  __gcd(int x,int y)
    {
        if(y==0)
            return x;
        return __gcd(y,x%y);
    }
    void push_up(int root)
    {
        if(tree[root<<1|1].value==tree[root<<1].value)
            tree[root].value=tree[root<<1].value;
        else
            tree[root].value=-1;
    }
    void push_down(int root)
    {
        if(tree[root].flag!=-1)
        {
            tree[root<<1].flag=tree[root<<1|1].flag=tree[root].flag;
            tree[root<<1].value=tree[root<<1|1].value=tree[root].flag;
            tree[root].flag=-1;
        }
    }
    void build(int root,int l,int r)
    {
        tree[root].l=l;
        tree[root].r=r;
        tree[root].flag=-1;
        if(l==r)
        {
            tree[root].value=a[r];
            return;
        }
        int mid=l+(r-l)/2;
        build(root<<1,l,mid);
        build(root<<1|1,mid+1,r);
        push_up(root);
    }
    void update1(int root,int l,int r,int x)//更改为某个值
    {
        if(l<=tree[root].l&&tree[root].r<=r)
        {
            tree[root].flag=tree[root].value=x;
            return;
        }
        push_down(root);
        int mid=tree[root].l+(tree[root].r-tree[root].l)/2;
        if(l<=mid)
            update1(root<<1,l,r,x);
        if(mid<r)
            update1(root<<1|1,l,r,x);
        push_up(root);
    }
    void update2(int root,int l,int r,int x)//变为gcd（）；
    {
        if(tree[root].value!=-1&&l<=tree[root].l&&tree[root].r<=r)
        {
            if(tree[root].value>x)
            {
                // printf("(%d %d)\n",x,__gcd(x,tree[root].value));
                tree[root].flag=tree[root].value=__gcd(tree[root].value,x);
            }
            return;
        }
        push_down(root);
        int mid=tree[root].l+(tree[root].r-tree[root].l)/2;
        if(l<=mid)
            update2(root<<1,l,r,x);
        if(mid<r)
            update2(root<<1|1,l,r,x);
        push_up(root);
    }
    void query(int root)
    {
        if(tree[root].l==tree[root].r)
        {
            a[tree[root].l]=tree[root].value;
            return;
        }
        push_down(root);
        query(root<<1);
        query(root<<1|1);
    }
} tr;
int main()
{
    int n,T,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        tr.build(1,1,n);
        scanf("%d",&m);
        while(m--)
        {
            int p,x,y,z;
            scanf("%d%d%d%d",&p,&x,&y,&z);
            if(p==1)
            {
                tr.update1(1,x,y,z);
            }
            else if(p==2)
            {
                tr.update2(1,x,y,z);
            }
        }
        tr.query(1);
        for(int i=1; i<=n; i++)
            printf(i==n?"%d \n":"%d ",a[i]);
    }
    return 0;
}