zoukankan      html  css  js  c++  java
  • HDU 1698 Just a Hook (线段树)

    Problem Description
    In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.



    Now Pudge wants to do some operations on the hook.

    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.
     
    Input
    The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.
    For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.
    Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.
     
    Output
    For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.
     
     
    Sample
    Sample Input
    1
    10
    2
    1 5 2
    5 9 3
     
    
    Sample Output
    Case 1: The total value of the hook is 24.

    题意:

      输入T,表示T组样例,输入n,表示有n个节点,最初所有点的值为1,然后给定X、Y、Z,将从X到Y改为权值为Z,计算权值和。

    思路:

      线段树区间更新的模板。

    代码:

    #include <cstdio>
    #include <string>
    #include <algorithm>
    #include <iostream>
    using namespace std;
    struct node
    {
        int l,r;
        int w;
    }tree[300005];
    void build(int l,int r,int n)
    {
        tree[n].l = l;
        tree[n].r = r;
        tree[n].w = 1;
        if(l == r)
            return ;
        int temp = (l+r)/2;
        build(l,temp,2*n);
        build(temp+1,r,2*n+1);
    }
    void update(int l,int r,int w,int n)
    {
        if(tree[n].l == l&&tree[n].r == r)
        {
            tree[n].w = w;
            return;
        }
        if(tree[n].w!=-1)//将值更改为-1,代表这个区间内的值都已经更改
        {
            tree[2*n].w = tree[2*n+1].w = tree[n].w;
            tree[n].w = -1;
        }
        int temp=(tree[n].l+tree[n].r)/2;
        if(r <= temp)
            update(l,r,w,2*n);
        else if(l>temp)
            update(l,r,w,2*n+1);
        else
        {
            update(l,temp,w,2*n);
            update(temp+1,r,w,2*n+1);
        }
    }
    int find(int n)//查找
    {
        if(tree[n].w != -1)//如果节点不为-1,表示这个区间内的值都是tree[n].w
        {
            return (tree[n].r-tree[n].l+1)*tree[n].w;
        }
        else//如果节点为-1,递归下去。
        {
            return find(2*n)+find(2*n+1);
        }
    }
    int main()
    {
        int T;
        scanf("%d",&T);
        for(int i=1; i<=T; i++)
        {
            int n;
            scanf("%d",&n);
            build(1,n,1);
            int p;
            scanf("%d",&p);
            while(p--)
            {
                int a,b,w;
                scanf("%d%d%d",&a,&b,&w);
                update(a,b,w,1);
            }
            printf("Case %d: ",i);
            printf("The total value of the hook is %d.
    ",find(1));
        }
    }
  • 相关阅读:
    简简单单制作鼠标静态动态 ani cur 小结 鼠标形状指针
    【VB6 学习文档管理系统源码】
    Delphi 中的全局快捷键+给指定窗体发送按键
    C# 委托实例实现的多种类型
    PyCharm 上传项目到码云托管平台
    vs rdlc 设置Tablix 在新页面重复表头
    .net C# Chart控件的简单使用
    发邮件,阿里云,未指定邮件服务器端口导致的报错
    使用Quartz Job 简单的做一个定时服务
    FromBase64String 输入的不是有效的 Base-64 字符串,因为它包含非 Base-64 字符、两个以上的填充字符,或者填充字符间包含非法字符
  • 原文地址:https://www.cnblogs.com/aiguona/p/7544935.html
Copyright © 2011-2022 走看看