【hdu】 Just a Hook （线段树 -成段更新，延迟标记）

Just a Hook

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 6   Accepted Submission(s) : 4

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 Input

1
10
2
1 5 2
5 9 3

 

Sample Output

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

 

这题花了不少时间，成段更新，延迟标记！
初始时，令所有tree[i].sum=1，此时，最上面一层即i=1时为可用的点。

每一步更新，不需要更新到底（单点更新），只需要更新tree[i].left==x&&tree[i].right==y即可，此点上面的均为不可用点！
最后查询，只需要找到可用点，（tree[i].right-tree[i].left+1)*tree[i].sum; 此时的sum必为更新后的值！
下面的代码很重要，当程序递归到第i个结点时，如果不为-1，则将它的左右儿子更新，将它置为不可用点，因为程序还要将递归到它的儿子节点。

1. if(tree[pos].sum!=-1)  
2.      {  
3.          tree[pos<<1].sum=tree[pos<<1|1].sum=tree[pos].sum;  
4.          tree[pos].sum=-1;  
5.      } 
   

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
struct segtree
{
       int left,right;
       int sum;
}tree[400005];
void Build(int pos,int x,int y)
{
     tree[pos].left=x;
     tree[pos].right=y;
     tree[pos].sum=1; 
     if(x==y){return ;}
     int mid=(x+y)>>1;
     Build((pos<<1),x,mid);
     Build((pos<<1|1),mid+1,y);
}
void Update(int pos,int x,int y,int z)
{
     if(tree[pos].sum==z)return; 
     int L=tree[pos].left;
     int R=tree[pos].right;
     int M=(L+R)>>1;
     if(L==x&&R==y) 
     {
         tree[pos].sum=z;
         return ;
     }
     if(tree[pos].sum!=-1)
     {
         tree[pos<<1].sum=tree[pos<<1|1].sum=tree[pos].sum;
         tree[pos].sum=-1;
     }
     if(y<=M) Update((pos<<1),x,y,z);  
     else if(x>M) Update(pos<<1|1,x,y,z);
     else
     { 
          Update((pos<<1),x,M,z); 
          Update(pos<<1|1,M+1,y,z);
     }
}
int Search(int i)
{
    if(tree[i].sum!=-1) return (tree[i].right-tree[i].left+1)*tree[i].sum;
    return Search(2*i)+Search(2*i+1);
}
int n;
int main()
{
    int T,i,j,t;
    scanf("%d",&T);
    for(i=1;i<=T;i++)
    {
              scanf("%d",&n);
              Build(1,1,n);
              scanf("%d",&t);
              int x,y,z;
              while(t--)
              {
                        scanf("%d%d%d",&x,&y,&z);
                        Update(1,x,y,z);
              }
              printf("Case %d: The total value of the hook is %d.\n",i,Search(1));
    }
    return 0;
}