【线段树】 hdu1698 Just a Hook

最新推荐文章于 2022-12-12 21:12:45 发布

转载最新推荐文章于 2022-12-12 21:12:45 发布 · 211 阅读

Just a Hook

Time Limit: 2000 MS

Memory Limit: 32768 K

Total Submit: 66 (43 users)

Total Accepted: 44 (43 users)

Special Judge: No

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.

https://blog.youkuaiyun.com/acm_ted/article/details/7786749

题意：图上这个胖子有根大棒子，这根大棒子由许多小棒子组成，小棒子初值为一，然后给你三个数，把对应区间的小棒棒换成相应的棒子

解法：注意这里是替换，而不是累加，即“置”（应该是这个字） 这里需要注意一下，因为要替换，所以得开个lazy标记，并且不用写查询函数，因为最后是问整个区间的和

让你修改[l, r]的类型为v, v的种类不同，分数不同，求所有操作后的总和，默认值是1

//区间段内的数进行整体的更新
#include<cstdio>
#include<cstring>
using namespace std;
#define MAX 100005
int sum[MAX<<2];
void push_up(int idx)
{
    sum[idx]=(sum[idx<<1]==sum[idx<<1|1])?sum[idx<<1]:-1;
}
void build(int l,int r,int idx)//建树
{
    if(l==r)
    {
        sum[idx]=1;
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,idx<<1);
    build(mid+1,r,idx<<1|1);
    push_up(idx);
    return;
}
void updata(int l,int r,int idx,int p,int a,int b)//更新
{
    if(a<=l&&r<=b)
    {
        sum[idx]=p;
        return;
    }
    if(sum[idx]>0)
        sum[idx<<1]=sum[idx<<1|1]=sum[idx];
    int mid=(l+r)>>1;
    if(a<=mid&&sum[idx<<1]!=p) updata(l,mid,idx<<1,p,a,b);
    if(b>mid&&sum[idx<<1|1]!=p) updata(mid+1,r,idx<<1|1,p,a,b);
    push_up(idx);
}
int query(int l,int r,int idx,int a,int b)//查询
{
    if(sum[idx]>0) return sum[idx]*(r-l+1);
    int mid=(l+r)>>1;
    int res=0;
    if(a<=mid) res+=query(l,mid,idx<<1,a,b);
    if(b>mid)  res+=query(mid+1,r,idx<<1|1,a,b);
    return res;
}
int main()
{
    int cas,n,q,a,b,p;
    scanf("%d",&cas);
    for(int t=1;t<=cas;++t)
    {
        scanf("%d%d",&n,&q);
        build(1,n,1);
        for(;q--;)
        {
            scanf("%d%d%d",&a,&b,&p);
            updata(1,n,1,p,a,b);
        }
        printf("Case %d: The total value of the hook is %d.\n",t,query(1,n,1,1,n));
    }
    return 0;
}