HDU 1698 Just a hook (线段树区间更新)

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#ACM #线段树 #区间更新

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本文介绍了一种解决Pudge's Hook问题的有效算法。通过使用线段树进行区间修改操作，实现了快速计算钩子总价值的功能。适用于大量更新操作场景。

Just a Hook

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31586 Accepted Submission(s): 15560

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

Sample Output

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

还是套用了我一直用的模板这题就不用写query啦!

上代码看注释吧!

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<stack>
#include<queue>
#include<cmath>
#include<stack>
#include<list>
#include<map>
#include<set>
typedef long long ll;
using namespace std;
struct node{
ll l,r,val;
ll add;
}nd[100005*4];
void build(ll l,ll r,ll k)
{
    nd[k].l=l;
    nd[k].r=r;
    nd[k].add=0;
    if(l==r)
    {
        nd[k].val=1;
        return;
    }
    ll mid=(l+r)/2;
    build(l,mid,2*k);
    build(mid+1,r,2*k+1);
    nd[k].val=nd[k*2].val+nd[k*2+1].val; //所谓的pushup()
}
void pushdown(ll k)
{
    ll t=2*k;
    nd[t].add=nd[k].add;  //注意 这次不是+=了,因为颜色会覆盖 所以要直接用=
    nd[t+1].add=nd[k].add;
    nd[t].val=nd[k].add*(nd[t].r-nd[t].l+1);
    nd[t+1].val=nd[k].add*(nd[t+1].r-nd[t+1].l+1);
    nd[k].add=0;
}
void update(ll l,ll r,ll c,ll k)
{
    if(nd[k].l>r||nd[k].r<l)   //这是这个模板的套路 在最开始判断区间有无交集
    {
        return;
    }
    if(nd[k].l>=l&&nd[k].r<=r)  //k节点所代表的区间完全在要更新的区间的时候 直接更新k结点
    {
        nd[k].add=c;
        nd[k].val=c*(nd[k].r-nd[k].l+1);
        return;
    }
    if(nd[k].add)
    {
        pushdown(k);
    }
    ll mid=(nd[k].r+nd[k].l)/2;
    if(r<=mid)update(l,r,c,2*k);
    else if(l>mid)update(l,r,c,2*k+1);
    else
    {
        update(l,mid,c,2*k);
        update(mid+1,r,c,2*k+1);
    }
    nd[k].val=nd[k*2].val+nd[k*2+1].val;
}



int main()
{
    int T;
    ll n,Q;
    ll x,y,z;
    while(scanf("%d",&T)==1)
    {
        int cnt=1;
        while(T--)
        {
            scanf("%lld",&n);
            build(1,n,1);
            scanf("%lld",&Q);
            while(Q--)
            {
                scanf("%lld%lld%lld",&x,&y,&z);
                update(x,y,z,1);
            }
            printf("Case %d: The total value of the hook is %lld.\n",cnt++,nd[1].val);
        }
    }

    return 0;
}