B. Light bulbs（2019 ICPC上海站）

題目描述：

There are NNN light bulbs indexed from 000 to N−1N-1N−1. Initially, all of them are off.A FLIP operation switches the state of a contiguous subset of bulbs. FLIP(L,R)FLIP(L, R)FLIP(L,R) means to flip all bulbs xxx such that L≤x≤RL \leq x \leq RL≤x≤R. So for example, FLIP(3,5)FLIP(3, 5)FLIP(3,5) means to flip bulbs 333 , 444 and 555, and FLIP(5,5)FLIP(5, 5)FLIP(5,5) means to flip bulb 555.Given the value of NNN and a sequence of MMM flips, count the number of light bulbs that will be on at the end state.

輸入：

The first line of the input gives the number of test cases, TTT. TTT test cases follow. Each test case starts with a line containing two integers NNN and MMM, the number of light bulbs and the number of operations, respectively. Then, there are MMM more lines, the iii-th of which contains the two integers LiL_iLi​ and RiR_iRi​, indicating that the iii-th operation would like to flip all the bulbs from LiL_iLi​ to RiR_iRi​ , inclusive.1≤T≤10001 \leq T \leq 10001≤T≤10001≤N≤1061 \leq N \leq 10^61≤N≤1061≤M≤10001 \leq M \leq 10001≤M≤10000≤Li≤Ri≤N−10 \leq L_i \leq R_i \leq N-10≤Li​≤Ri​≤N−1

輸出：

For each test case, output one line containing Case #x: y, where xxx is the test case number (starting from 111) and yyy is the number of light bulbs that will be on at the end state, as described above.

樣例輸入：

2
10 2
2 6
4 8
6 3
1 1
2 3
3 4

樣例輸出：

Case #1: 4
Case #2: 3

code：

這道題，當時寫的時候，實在是一言難盡，一開始感覺是線段樹？？然後寫了寫發現超內存，然後換成樹狀數組？？然後寫了寫發現超時，超時在分個塊查詢？？？然後怎麼分都是超時，超時，超時，，，，自閉。。。。。
代碼其實很簡單，初學者都能看懂，只是這個思路，有點難想

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=1e6+20;
int num[maxn*2];
int main()
{
 int ttt;
 scanf("%d",&ttt);
 for(int kk=1;kk<=ttt;kk++)
 {
  int n,m;
  scanf("%d%d",&n,&m);
  int count=0;
  for(int i=0;i<m;i++)
  {
   int xx,yy;
   scanf("%d%d",&xx,&yy);
   num[count++]=xx;
   num[count++]=yy+1;
  }
  sort(num,num+m*2);
  int sum=0;
  for(int i=0;i<count;i+=2)
  {
   sum+=num[i+1]-num[i];
  }
  printf("Case #%d: %d\n",kk,sum);
 }
 return 0;
 }