Just a Hook(线段树区间修改值)-------------蓝桥备战系列-优快云博客

本文介绍了一种算法，用于计算游戏DotA中角色Pudge的MeatHook在进行一系列金属棒替换操作后的总价值。通过使用线段树和懒惰传播优化，解决了大量更新和查询操作的问题。

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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

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.\n",cnt,query(1,1,n));
	  cnt++;
	}

	return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/10781755.html