# Hdu1698just a hook

Posted by yjjr's blog on February 6, 2018

HDU1698 Just a Hook

http://www.sakurasake.icoc.me/nd.jsp?id=6#_np=2_327

Time Limit: 4000/2000 MS(Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

In the game of DotA, Pudge’s meat hook isactually the most horrible thing for most of the heroes. The hook is made up ofseveral 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 eachoperation, Pudge can change the consecutive metallic sticks, numbered from X toY, into cupreous sticks, silver sticks or golden sticks.
The total value of the hook is calculated as the sum of values of N metallicsticks. More precisely, the value for each kind of stick is calculated asfollows:

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 theoperations.
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 than10 cases.
For each case, the first line contains an integer N, 1<=N<=100,000, whichis the number of the sticks of Pudge’s meat hook and the second line containsan 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 toY into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 representsthe silver kind and Z=3 represents the golden kind.

Output

For each case, print a number in a linerepresenting the total value of the hook after the operations. Use the formatin 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.

Source

#include<iostream>
#include<iomanip>
#include<cmath>
#include<cstring>
#include<ctime>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
#define N 100001
{
int f=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
using namespace std;
int st[N<<2];
int flag;

void make(int l,int r,int k)
{
st[k]=flag;
if(l==r)return;
int m=(l+r)>>1;
make(lson);
make(rson);
}

void down(int &k)
{
st[k<<1]=st[k<<1|1]=st[k];
st[k]=0;
}

void change(int &L,int &R,int l,int r,int k)
{
if(L<=l&&R>=r)
{
st[k]=flag;
return;
}//如果寻找需要更新的线段正好处于某个区间内，那么一整段都更新
if(st[k])down(k);//如果该节点为纯色，那么down操作，左右儿子颜色更新为纯色
int m=(l+r)>>1;
if(L<=m)change(L,R,lson);
if(R>m)change(L,R,rson);//分别递归更新左右儿子
}

int query(int l,int r,int k)
{
if(st[k])
return st[k]*(r-l+1);
int m=(l+r)>>1,t1,t2;
t1=query(lson);
t2=query(rson);
return t1+t2;
}

int main()
{
int T,n,t=1;
int L,R,q;
while(T--)
{
flag=1;
make(1,n,1);
while(q--)
{