Just a Hook

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 67   Accepted Submission(s) : 28

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

11021 5 25 9 3

 

Sample Output

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

 

Source

2008 “Sunline Cup” National Invitational Contest

     题意：给n个钩子一开始全是银，然后给出操作数，将区间[x,y]上的钩子类型改为z类型，z=1，2，3，金银铜，求最后所有钩子价值

     区间更新，区间查询

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define maxn 100005
int tree[maxn<<2],lazy[maxn<<2];//求区间和，延迟标记
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
void pushup(int rt)//向上更新
{
    tree[rt]=tree[rt<<1]+tree[rt<<1|1];
}
void pushdown(int rt,int m)//
{
    if(lazy[rt])//往下更新
    {
        lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt];
        tree[rt<<1]=(m-(m>>1))*lazy[rt];
        tree[rt<<1|1]=(m>>1)*lazy[rt];
        lazy[rt]=0;//往下更新完后，标记为为0，当前不需往下更新
    }
}
void build(int l,int r,int rt)
{
    lazy[rt]=0;
    //tree[rt]=r-l+1;
    int m=(l+r)>>1;
    if(l==r)
    {tree[rt]=1;return ;}
    build(lson);
    build(rson);
    pushup(rt);
}

void update(int x,int y,int z,int l,int r,int rt)//区间更新
{
    if(x<=l&&r<=y)
    {
        tree[rt]=z*(r-l+1);
        lazy[rt]=z;
        return ;
    }
    pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    if(x<=m)
       update(x,y,z,lson);
    if(y>m)
        update(x,y,z,rson);
    pushup(rt);
}
int main()
{
    int t,n,q,x,y,z,i;
    scanf("%d",&t);
    for(i=1;i<=t;i++)
    {
        scanf("%d%d",&n,&q);
        build(1,n,1);
        while(q--)
        {
            scanf("%d%d%d",&x,&y,&z);
            update(x,y,z,1,n,1);
        }
        cout<<"Case "<<i<<": The total value of the hook is "<<tree[1]<<"."<<endl;//求得是整个区间的和
    }
}