hdu 1698 Just a Hook (线段树区间更新+懒惰标记)

最新推荐文章于 2019-09-14 10:51:01 发布

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#hdu #线段树 #懒惰标记

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解决了一个关于游戏《DotA》中角色Pudge的钩子问题，通过使用懒惰标记优化线段树算法来高效更新和计算钩子的价值。

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

Sample Output

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

分析：如果用朴素的线段树更新到叶子的话会超时，这里采用懒惰标记，如果更新到某一个区间，先把这个区间标记上，然后等下次更新经过这个区间的时候，再来更新这个节点的左右儿子并标记。这样就可以减少大量的时间。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=100005;
int tr[4*N];
int add[4*N];
void PushUp(int i)
{
    tr[i]=tr[2*i]+tr[2*i+1];
}
void PushDown(int i,int len)
{
    if(add[i])
    {
        add[2*i]=add[2*i+1]=add[i];
        tr[i<<1]=(len-(len>>1))*add[i];
        tr[i<<1|1]=(len>>1)*add[i];
        add[i]=0;
    }
}
void build(int i,int l,int r)
{
    tr[i]=1;
    add[i]=0;
    if(l==r)
    {
        return;
    }
    int mid=(l+r)/2;
    build(2*i,l,mid);
    build(2*i+1,mid+1,r);
    PushUp(i);
}
void update(int i,int l,int r,int x,int y,int c)
{
    if(x<=l&&r<=y)
    {
        add[i]=c;
        tr[i]=(r-l+1)*c;
        return;
    }
    PushDown(i,r-l+1);
    int mid=(l+r)/2;
    if(x<=mid) update(2*i,l,mid,x,y,c);
    if(y>mid) update(2*i+1,mid+1,r,x,y,c);
    PushUp(i);
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int k=1;k<=t;k++)
    {
        int n,times,l,r,c;
        scanf("%d%d",&n,×);
        build(1,1,n);
        for(int i=1;i<=times;i++)
        {
            scanf("%d%d%d",&l,&r,&c);
            update(1,1,n,l,r,c);
        }
        printf("Case %d: The total value of the hook is %d.\n",k,tr[1]);

    }
    return 0;
}