Just a Hook---线段树--区间赋值--区间查询

我就是我 2023-07-23 11:56 2阅读 0赞

## 题目描述 ##

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.  
![在这里插入图片描述][format_png]  
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.

## 题目解法 ##

大水题一枚，这题就是套线段树模板，将pushdowm的操作改成赋值操作即可，还要记得初始化懒人标记。  
代码如下：

#include<algorithm>
    #include<cstdio>
    using namespace std;
    #define ll long long
    inline int read()
    {
        
        int x=0,sign=1;
        char ch;
        ch=getchar();
        while(ch<'0'||ch>'9')
        {
        
            if(ch=='-')
                sign=-1;
            ch=getchar();
        }
        while(ch>='0'&&ch<='9')
        {
        
            x=(x<<1)+(x<<3)+(ch^48);
            ch=getchar();
        }
        return sign*x;
    }
    typedef struct
    {
        
        int l,r,w,lazy=0;
    } NODE;
    NODE tree[400005];
    void build(int x,int y,int k)
    {
        
        tree[k].lazy=0;
        tree[k].l=x;
        tree[k].r=y;
        if(x==y)
        {
        
            tree[k].w=1;
            return ;
        }
        int mid=x+(y-x)/2;
        build(x,mid,k<<1);
        build(mid+1,y,k<<1|1);
        tree[k].w=tree[k<<1].w+tree[k<<1|1].w;
    }
    void pushdowm(int k)
    {
        
        if(tree[k].l==tree[k].r)
        {
        
            tree[k].lazy=0;
            return ;
        }
        tree[k<<1].w=(tree[k<<1].r-tree[k<<1].l+1)*tree[k].lazy;
        tree[k<<1|1].w=(tree[k<<1|1].r-tree[k<<1|1].l+1)*tree[k].lazy;
        tree[k<<1].lazy=tree[k].lazy;
        tree[k<<1|1].lazy=tree[k].lazy;
        tree[k].lazy=0;
    }
    void change(int x,int y,int t,int k)
    {
        
        if(tree[k].lazy)
            pushdowm(k);
        if(tree[k].l==x&&tree[k].r==y)
        {
        
            tree[k].w=(tree[k].r-tree[k].l+1)*t;
            tree[k].lazy=t;
            return;
        }
        int mid=tree[k].l+(tree[k].r-tree[k].l)/2;
        if(mid>=y)
            change(x,y,t,k<<1);
        else if(mid<x)
            change(x,y,t,k<<1|1);
        else
        {
        
            change(x,mid,t,k<<1);
            change(mid+1,y,t,k<<1|1);
        }
        tree[k].w=tree[k<<1].w+tree[k<<1|1].w;
    }
    int ask_query(int x,int y,int k)
    {
        
        if(tree[k].lazy)
            pushdowm(k);
        if(tree[k].l==x&&tree[k].r==y)
        {
        
            return tree[k].w;
        }
        int mid;
        mid=tree[k].l+(tree[k].r-tree[k].l)/2;
        if(mid>=y)
            return ask_query(x,y,k<<1);
        else if(mid<x)
            return ask_query(x,y,k<<1|1);
        return ask_query(x,mid,k<<1)+ask_query(mid+1,y,k<<1|1);
    }
    int main()
    {
        
        int T;
        T=read();
        for(int i=1; i<=T; i++)
        {
        
            int n,m;
            n=read();
            m=read();
            build(1,n,1);
            for(int j=0; j<m; j++)
            {
        
                int a,b,c;
                a=read();
                b=read();
                c=read();
                change(a,b,c,1);
            }
            int ans;
            ans=ask_query(1,n,1);
            printf("Case %d: The total value of the hook is %d.\n",i,ans);
        }
        return 0;
    }

[format_png]: /images/20230528/20cec590c06243d08fb63672dcb7f29a.png