HDU - 4027 Can you answer these queries? （线段树区间更新区间查询）

最新推荐文章于 2021-11-15 22:13:06 发布

xuejye

最新推荐文章于 2021-11-15 22:13:06 发布

阅读量189

点赞数

CC 4.0 BY-SA版权

分类专栏：树状数组&&线段树

本文链接：https://blog.youkuaiyun.com/xuejye/article/details/77203440

树状数组&&线段树专栏收录该内容

32 篇文章

订阅专栏

本文介绍了一个涉及区间更新和区间查询的算法问题，通过构建线段树实现对一系列战舰耐久值的高效处理。每艘战舰有一个初始耐久值，武器攻击会将指定范围内战舰的耐久值更新为其平方根（向下取整），并能查询任意区间的总耐久值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4027点击打开链接

Can you answer these queries?

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 18426 Accepted Submission(s): 4344

Problem Description

A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon, it could decrease the endurance of a consecutive part of battleships by make their endurance to the square root of it original value of endurance. During the series of attack of our secret weapon, the commander wants to evaluate the effect of the weapon, so he asks you for help.
You are asked to answer the queries that the sum of the endurance of a consecutive part of the battleship line.

Notice that the square root operation should be rounded down to integer.

Input

The input contains several test cases, terminated by EOF.
  For each test case, the first line contains a single integer N, denoting there are N battleships of evil in a line. (1 <= N <= 100000)
  The second line contains N integers Ei, indicating the endurance value of each battleship from the beginning of the line to the end. You can assume that the sum of all endurance value is less than 2 ⁶³.
  The next line contains an integer M, denoting the number of actions and queries. (1 <= M <= 100000)
  For the following M lines, each line contains three integers T, X and Y. The T=0 denoting the action of the secret weapon, which will decrease the endurance value of the battleships between the X-th and Y-th battleship, inclusive. The T=1 denoting the query of the commander which ask for the sum of the endurance value of the battleship between X-th and Y-th, inclusive.

Output

For each test case, print the case number at the first line. Then print one line for each query. And remember follow a blank line after each test case.

Sample Input

  
   10
1 2 3 4 5 6 7 8 9 10
5
0 1 10
1 1 10
1 1 5
0 5 8
1 4 8

Sample Output

  
   Case #1:
19
7
6

区间更新区间查询操作

poj的g++用longlong c++用__int64

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define maxn 1000000+5
struct xjy
{
    int lazy;
    int left;
    int right;
    __int64 sum;
};
xjy tree[maxn<<2];
__int64 a[maxn];
__int64 aans;
void show(int i,int left,int right)
{
    if(left==right)
    {
        printf("%I64d %I64d %I64d\n",tree[i].left,tree[i].right,tree[i].sum);
        return ;
    }
    int mid=(left+right)>>1;
    show(i<<1,left,mid);
    show(i<<1|1,mid+1,right);
    printf("%I64d %I64d %I64d\n",tree[i].left,tree[i].right,tree[i].sum);
}
void build(int i,int left ,int right)
{
    if(left==right)
    {
        tree[i].left=left;
        tree[i].right=right;
        tree[i].sum=a[left];
        tree[i].lazy=1;
        if(a[left]<=1)
            tree[i].lazy=0;
        return ;
    }
    int mid=(left+right)>>1;
    build(i<<1,left,mid);
    build (i<<1|1,mid+1,right);
    tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;
    tree[i].left=left;
    tree[i].right=right;
}

void updateregion(int i,int left,int right)
{
    if(!tree[i].lazy)
    {
        return;
    }
    if(tree[i].left==left&&tree[i].right==right&&tree[i].left==tree[i].right)
    {
        tree[i].sum=(__int64)sqrt(1.0*tree[i].sum);
        if(tree[i].sum==1)
            tree[i].lazy=0;
        return;
    }
    int mid=(tree[i].left+tree[i].right)>>1;
    if(right<=mid)
        updateregion(i<<1,left,right);
    else if(left>mid)
        updateregion(i<<1|1,left,right);
    else
    {
        updateregion(i<<1,left,mid);
        updateregion(i<<1|1,mid+1,right);
    }
    tree[i].sum=tree[i<<1].sum+tree[i<<1|1].sum;
    if(tree[i].sum==(tree[i].right-tree[i].left+1))
        tree[i].lazy=0;
}

void queryregion(int i,int left,int right)
{
    if(left==tree[i].left&&right==tree[i].right)
    {
        aans+=tree[i].sum;
        return ;
    }
    int mid=(tree[i].left+tree[i].right)>>1;
    if(right<=mid)
        queryregion(i<<1,left,right);
    else if(left>mid)
        queryregion(i<<1|1,left,right);
    else
        {
            queryregion(i<<1,left,mid);
            queryregion(i<<1|1,mid+1,right);
        }
}
int main()
{
    __int64 n,m;
    __int64 t=1;
    while(~scanf("%I64d",&n))
        {
            printf("Case #%I64d:\n",t++);
            for(int i=1;i<maxn*4;i++)
            {
                tree[i].lazy=1;
                tree[i].left=0;
                tree[i].right=0;
                tree[i].sum=0;
                a[i]=0;
            }
            for(int i=1;i<=n;i++)
                scanf("%I64d",&a[i]);
            build(1,1,n);
            //show(1,1,n);
            scanf("%I64d",&m);
            for(int i=1;i<=m;i++)
            {
                int l,r,type;
                scanf("%d%d%d",&type,&l,&r);
                if(l>r)
                    swap(l,r);
                    else ;
                if(type==0)
                    updateregion(1,l,r);//show(1,1,n);
                else
                {
                    //show(1,1,n);
                    aans=0;
                    queryregion(1,l,r);
                    printf("%I64d\n",aans);
                }
            }
            printf("\n");
        }

}