zoukankan      html  css  js  c++  java
  • Just a Hook(线段树区间修改值)-------------蓝桥备战系列

    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 Input

    1
    10
    2
    1 5 2
    5 9 3

    Sample Output

    Case 1: The total value of the hook is 24.

    注意后面这个标点‘.’

    代码:
     

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #include<queue>
    #include<stack>
    #include<map>
    #include<set>
    #include<vector>
    #include<cmath>
    
    const int maxn=1e5+5;
    typedef long long ll;
    using namespace std;
    
    struct node
    {
    	ll l,r,sum;
    }tree[maxn<<2];
    ll lazy[maxn<<2];
    void pushup(int m)
    {
    
    	tree[m].sum=tree[m<<1].sum+tree[m<<1|1].sum;
    }
    void pushdown(int m,int l)
    {
    	if(lazy[m]!=0)
    	{
    		lazy[m<<1]=lazy[m];
    		lazy[m<<1|1]=lazy[m];
    		tree[m<<1].sum=lazy[m]*(l-(l>>1));
    		tree[m<<1|1].sum=lazy[m]*(l>>1);
    		lazy[m]=0;
    	}
    }
    void build(int m,int l,int r)
    {
    	tree[m].l=l;
    	tree[m].r=r;
    	lazy[m]=0;
    	if(l==r)
    	{
    	    tree[m].sum=1;
    		return;
    	}
    	int mid=(l+r)>>1;
    	build(m<<1,l,mid);
    	build(m<<1|1,mid+1,r);
    	pushup(m);
    }
    void update(int m,int l,int r,int val)
    {
    	if(tree[m].l==l&&tree[m].r==r)
    	{
    		lazy[m]=val;
    		tree[m].sum=(ll)val*(r-l+1);
    		return;
    	}
    	if(tree[m].l==tree[m].r)
    	return;
    	int mid=(tree[m].l+tree[m].r)>>1;
    	pushdown(m,tree[m].r-tree[m].l+1);
    	if(r<=mid)
    	{
    		update(m<<1,l,r,val);
    	}
    	else if(l>mid)
    	{
    		update(m<<1|1,l,r,val);
    	}
    	else 
    	{
    		update(m<<1,l,mid,val);
    		update(m<<1|1,mid+1,r,val);
    	}
    	pushup(m);
    }
    ll query(int m,int l,int r)
    {
    	if(tree[m].l==l&&tree[m].r==r)
    	{
    		return tree[m].sum;
    	}
    	pushdown(m,tree[m].r-tree[m].l+1);
    	int mid=(tree[m].l+tree[m].r)>>1;
    	ll res=0;
    	if(r<=mid)
    	{
    		res+=query(m<<1,l,r);
    	}
    	else if(l>mid)
    	{
    		res+=query(m<<1|1,l,r);
    	}
    	else 
    	{
    		res+=(query(m<<1,l,mid)+query(m<<1|1,mid+1,r));
    		
    	}
    	return res;
    	
    }
    int main()
    {
        int T;
        cin>>T;
        int n,m;
        int cnt=1;
        while(T--)
        {
    	  cin>>n;
    	  build(1,1,n);
    	  cin>>m;
    	  char op[2];
    	  int l,r,val;
    	  for(int t=0;t<m;t++)
    	  {
    	  	scanf("%d%d%d",&l,&r,&val);
    	  	update(1,l,r,val);
    	  }
    	  printf("Case %d: The total value of the hook is %lld.
    ",cnt,query(1,1,n));
    	  cnt++;
    	}
    
    	return 0;
    }
    
  • 相关阅读:
    【解决火车轮播图小圆点跳的问题】传统轮播图-三位法
    jq龙禧轮播图
    QT MSVC环境中添加QWT
    XDMA ip core的使用
    PCIe基础知识与例程分析----PIO_demo
    Day04 (四)_TCP文件传输设计
    Day04 (三)_UDP传输端设计
    Day04 (二)_TCP传输客户器端设计
    Day04 (一)_TCP传输服务器端设计
    Day03 (下)_Qt文件系统
  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10781755.html
Copyright © 2011-2022 走看看