K - Just a Hook

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.            
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
#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;#define MAXN 100010struct node{    int l,r,sum,sign,lazy;//lazy标记，sign记录值，不过我这里貌似逗逼了，sign，lazy一个就够了。。够了。。了。。}tr[MAXN*4];void build(int id,int l,int r){    tr[id].l=l;    tr[id].r=r;    tr[id].sign=0;//区间更新初始化    tr[id].lazy=0;    if(l==r)        tr[id].sum=1;    else    {        int mid=(l+r)/2;        build(id*2,l,mid);        build(id*2+1,mid+1,r);        tr[id].sum=tr[id*2].sum+tr[id*2+1].sum;    }}void add(int id,int l,int r,int val){    if(l==tr[id].l&&tr[id].r==r)    {        tr[id].lazy=1;        tr[id].sign=val;        tr[id].sum=(r-l+1)*val;//注意此处求和啊    }    else    {        int mid=(tr[id].l+tr[id].r)/2;        if(tr[id].lazy==1)//判断是否应该向下更新，额。。其实这里一个lazy就搞定了        {            add(id*2,tr[id].l,mid,tr[id].sign);            add(id*2+1,mid+1,tr[id].r,tr[id].sign);            tr[id].sign=0;            tr[id].lazy=0;        }        if(r<=mid)//全在左区间            add(id*2,l,r,val);        else if (l>mid)//全在右区间            add(id*2+1,l,r,val);        else//两边都有        {            add(id*2,l,mid,val);            add(id*2+1,mid+1,r,val);        }        tr[id].sum=tr[id*2].sum+tr[id*2+1].sum;//别忘了求和    }}int main(){    int m,n,i,k,x,y,z,t=0;    scanf("%d",&m);    while(m--)    {        t++;        scanf("%d",&k);        build(1,1,k);        scanf("%d",&n);        while(n--)        {            scanf("%d%d%d",&x,&y,&z);            add(1,x,y,z);        }        printf("Case %d: The total value of the hook is %d.\n",t,tr[1].sum);//输出tr[1].sum即可    }    return 0;}