Just a Hook

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

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.

  
  

   
   1
10
2
1 5 2
5 9 3
  
  

  
  

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

#include<stdio.h>
#define NN 250000
struct node
{
	int l,r,ans,flag;
}g[NN*3];
void build_tree(int l,int r,int h)
{
	g[h].l=l;g[h].r=r;g[h].flag=0;
	if(l==r)
	{
		g[h].ans=1;
		return ;		
	}
	int mid=(g[h].l+g[h].r)/2;
	build_tree(l,mid,h+h);
	build_tree(mid+1,r,h+h+1);
	g[h].ans=g[h+h].ans+g[h+h+1].ans;
}
void update(int l,int r,int h,int add)
{
	if(l==g[h].l&&g[h].r==r)
	{
		g[h].flag=add;
		g[h].ans=(g[h].r-g[h].l+1)*g[h].flag;
		return;
	}
	if(g[h].flag)
	{
		g[h+h].flag=g[h+h+1].flag=g[h].flag;
		g[h+h].ans=(g[h+h].r-g[h+h].l+1)*g[h].flag;
		g[h+h+1].ans=(g[h+h+1].r-g[h+h+1].l+1)*g[h].flag;
		g[h].flag=0;
	}
	int mid=(g[h].l+g[h].r)/2;
	if(r<=mid)
	update(l,r,h+h,add);
	else if(l>mid)
	update(l,r,h+h+1,add);
	else
	{
		update(l,mid,h+h,add);
		update(mid+1,r,h+h+1,add);
	}
	g[h].ans=g[h+h].ans+g[h+h+1].ans;	
}
int quire(int l,int r,int h)
{
	if(l==g[h].l&&r==g[h].r)
	{
		return g[h].ans;
	}
	if(g[h].flag)
	{
		g[h+h].flag=g[h+h+1].flag=g[h].flag;
		g[h+h].ans=(g[h+h].r-g[h+h].l+1)*g[h].flag;
		g[h+h+1].ans=(g[h+h+1].r-g[h+h+1].l+1)*g[h].flag;
		g[h].flag=0;
	}
	int mid=(g[h].l+g[h].r)/2;
	if(r<=mid)
	return quire(l,r,h+h);
	else if(l>mid)
	return quire(l,r,h+h+1);
	else 
	return quire(l,mid,h+h)+quire(mid+1,r,h+h+1);
}
int main()
{
	int i,j,k,m,n,a,b,c,ncase,ww;
	scanf("%d",&ncase);
	for(ww=1;ww<=ncase;ww++)
	{
		scanf("%d%d",&m,&n);
	    build_tree(1,m,1);
	    while(n--)
	    {
    		scanf("%d%d%d",&a,&b,&c);
    		update(a,b,1,c);
    	}
    	printf("Case %d: The total value of the hook is %d.\n",ww,quire(1,m,1));
	}
	return 0;
}