线段树区间更新——HDU 1698

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Just a Hook

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

Submit Status

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.
        
     
     
     
     

 

Source

题意：Q个询问，每个询问(x,y,z）表示x~y的数变成z，求所有询问后1~n的总和

思路：线段树区间更新模板题，就是使用结构体跟不使用结构体要注意下~没什么可说。。。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<iostream>
#define ms(x,y) memset(x,y,sizeof(x))
const int MAXN=100000+10;
const int INF=1<<30;
using namespace std;
int sum[4*MAXN];
int add[4*MAXN];
int res;

void down(int rt, int len)
{
	if(add[rt])
	{
		int l=rt<<1, r=rt<<1|1, num=add[rt];
		add[l]=num;
		add[r]=num;
		sum[l]=num*(len-(len>>1));
		sum[r]=num*(len>>1);
		add[rt]=0;
	}
}

void buildtree(int rt, int left, int right)
{
	if(left==right){
		sum[rt]=1;
		return;
	}
	int mid=(left+right)>>1;
	buildtree(rt<<1, left, mid);
	buildtree(rt<<1|1, mid+1, right);
	sum[rt]=sum[rt<<1] + sum[rt<<1|1];
}

void updata(int rt, int left, int right, int x, int y, int z)
{
	if(x==left && right==y){
		add[rt]=z;
		sum[rt]=z*(right-left+1);
		return;
	}
	down(rt, right-left+1);
	int mid=(left+right)>>1;
	if(mid>=y) updata(rt<<1, left, mid, x, y, z);
	else if(mid<x) updata(rt<<1|1, mid+1, right, x, y, z);
	else{
		updata(rt<<1, left, mid, x, mid, z);
		updata(rt<<1|1, mid+1, right, mid+1, y, z);
	}
	sum[rt]=sum[rt<<1] + sum[rt<<1|1];
}

int main()
{
#if 0
	freopen("in.txt","r",stdin);
#endif
	int T,w=0;
	scanf("%d", &T);
	while(T--)
	{
		ms(add,0);
		res=0;
		int n,q;
		scanf("%d%d", &n,&q);
		buildtree(1,1,n);
		for(int i=0; i<q; i++){
			int x,y,z;
			scanf("%d%d%d", &x,&y,&z);
			updata(1,1,n,x,y,z);
		}
		printf("Case %d: The total value of the hook is %d.\n", ++w,sum[1]);
	}
	return 0;
}