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HDU4902-Nice boat（线段树）

2024-07-17 15:39:07   221人阅读

Nice boat

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1455    Accepted Submission(s): 645


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
 

Author
WJMZBMR
 

题意：n个数，Q组操作： 一、将[L,R] 的数设成X， 二、将[L,R]中大于X的数设成与X的GCD。最后输出所有的数
思路：线段树，标记线段，pushdown 和 pushup的使用，主要的一个优化的地方就是添加一个区间最大值，如果最大值小于X那么不更新整个区间。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
const int maxn = 100000+10;
int n,q;
int num[maxn];
struct node{
    int lson,rson;
    int maxx;
    int num;
    bool flag;
    int mid(){
        return (lson+rson)>>1;
    }
}tree[maxn*4];
int gcd(int a,int b){
    if(b==0) return a;
    return gcd(b,a%b);
}
void pushUp(int rt){
    tree[rt].maxx = max(tree[rt<<1].maxx,tree[rt<<1|1].maxx);
}
void pushDown(int rt){
    if(tree[rt].flag){
        tree[rt<<1].maxx = tree[rt<<1|1].maxx = tree[rt].maxx;
        tree[rt<<1].flag = tree[rt<<1|1].flag = true;
        tree[rt].flag = false;
    }
}
void build(int L,int R,int rt){
    tree[rt].lson = L;
    tree[rt].rson = R;
    tree[rt].flag = false;
    if(L==R){
        tree[rt].maxx = num[L];
        tree[rt].flag = true;
        return;
    }
    int mid = tree[rt].mid();
    build(L,mid,rt<<1);
    build(mid+1,R,rt<<1|1);
    pushUp(rt);
}
void update1(int L,int R,int l,int r,int rt,int x){
    if(L <= l && R >= r){
        tree[rt].flag = true;
        tree[rt].maxx = x;
        return;
    }
    pushDown(rt);
    int mid = tree[rt].mid();
    if(L <= mid){
        update1(L,R,l,mid,rt<<1,x);
    }
    if(R > mid){
        update1(L,R,mid+1,r,rt<<1|1,x);
    }
    pushUp(rt);
}
void update2(int L,int R,int l,int r,int rt,int x){
    if(tree[rt].maxx < x) return;
    if(L <= l && R >= r && tree[rt].flag && tree[rt].maxx > x){
        tree[rt].maxx = gcd(tree[rt].maxx,x);
        return;
    }
    pushDown(rt);
    int mid = tree[rt].mid();
    if(L<=mid) update2(L,R,l,mid,rt<<1,x);
    if(R>mid) update2(L,R,mid+1,r,rt<<1|1,x);
    pushUp(rt);
}
void print(int L,int R,int rt){
    if(L==R){
        printf("%d ",tree[rt].maxx);
        return;
    }
    pushDown(rt);
    int mid = tree[rt].mid();
    print(L,mid,rt<<1);
    print(mid+1,R,rt<<1|1);
}
int main(){

    int ncase;
    cin >> ncase;
    while(ncase--){
        scanf("%d",&n);
        for(int i = 1; i <= n; i++) scanf("%d",&num[i]);
        build(1,n,1);
        int a,b,c,d;
        scanf("%d",&q);
        while(q--){
            scanf("%d%d%d%d",&a,&b,&c,&d);
            if(a==1){
                update1(b,c,1,n,1,d);
            }else{
                update2(b,c,1,n,1,d);
            }
        }
        print(1,n,1);
        printf("\n");
    }
    return 0;
}



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