zoukankan      html  css  js  c++  java
  • hdu 1698 Just a Hook 线段树区间更新

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1698

    Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
    The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:

    For each cupreous stick, the value is 1.
    For each silver stick, the value is 2.
    For each golden stick, the value is 3.

    Pudge wants to know the total value of the hook after performing the operations.
    You may consider the original hook is made up of cupreous sticks.

    题意描述:给出n个数,初始化为1,然后Q个操作,每个操作x,y,z如下:

    z为1:更新从第x个数到第y个数连续区间为1.

    z为2:更新从第x个数到第y个数连续区间为2.

    z为3:更新从第x个数到第y个数连续区间为3.

    最后求出这个n个数的和。

    算法分析:线段树区间更改的做法,设立sum[]和col[](lazy标记)。

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstdlib>
     4 #include<cstring>
     5 #include<cmath>
     6 #include<algorithm>
     7 #define inf 0x7fffffff
     8 using namespace std;
     9 const int maxn=100000+10;
    10 
    11 int n,q;
    12 int sum[maxn<<2],col[maxn<<2];
    13 
    14 void PushUP(int rt) {sum[rt]=sum[rt<<1]+sum[rt<<1|1]; }
    15 void PushDown(int rt,int m)
    16 {
    17     if (col[rt]!=-1)
    18     {
    19         col[rt<<1]=col[rt<<1|1]=col[rt];
    20         sum[rt<<1]=(m-m/2)*col[rt];
    21         sum[rt<<1|1]=(m/2)*col[rt];
    22         col[rt]=-1;
    23     }
    24 }
    25 
    26 void build(int l,int r,int rt)
    27 {
    28     col[rt]=-1;
    29     if (l==r)
    30     {
    31         sum[rt]=1;
    32         return;
    33     }
    34     int m=(l+r)>>1;
    35     build(l,m,rt<<1);
    36     build(m+1,r,rt<<1|1);
    37     PushUP(rt);
    38 }
    39 
    40 void update(int l,int r,int rt,int x,int y,int z)
    41 {
    42     if (x<=l && r<=y)
    43     {
    44         col[rt]=z ;sum[rt]=(r-l+1)*z;
    45         return;
    46     }
    47     int mid=(l+r)>>1;
    48     PushDown(rt,r-l+1);
    49     if (y<=mid) update(l,mid,rt<<1,x,y,z);
    50     else if (x>mid) update(mid+1,r,rt<<1|1,x,y,z);
    51     else {update(l,mid,rt<<1,x,mid,z);update(mid+1,r,rt<<1|1,mid+1,y,z); }
    52     PushUP(rt);
    53 }
    54 
    55 void query(int l,int r,int rt)
    56 {
    57     if (l==r) return;
    58     PushDown(rt,r-l+1);
    59     int mid=(l+r)>>1;
    60     query(l,mid,rt<<1);
    61     query(mid+1,r,rt<<1|1);
    62     PushUP(rt);
    63 }
    64 
    65 int main()
    66 {
    67     int t,ncase=1;scanf("%d",&t);
    68     while (t--)
    69     {
    70         scanf("%d%d",&n,&q);
    71         int x,y,z;
    72         build(1,n,1);
    73         //cout<<sum[1]<<endl;
    74         for (int i=1 ;i<=q ;i++)
    75         {
    76             scanf("%d%d%d",&x,&y,&z);
    77             update(1,n,1,x,y,z);
    78         }
    79         query(1,n,1);
    80         printf("Case %d: The total value of the hook is %d.
    ",ncase++,sum[1]);
    81     }
    82     return 0;
    83 }
  • 相关阅读:
    zabbix获取主从同步状态的shell
    Linux 线程占用CPU过高定位分析
    LVM
    zabbix添加端口连接数监控
    hadoop未授权访问漏洞
    nginx修改上传文件大小限制(413 Request Entity Too Large)
    官方Tomcat 7.0.81 漏洞整改记录
    python流程控制:while循环
    python流程控制:for循环
    python数据类型—列表(增改删查,统计,取值,排序)
  • 原文地址:https://www.cnblogs.com/huangxf/p/4391145.html
Copyright © 2011-2022 走看看