B Light bulbs （2019icpc上海网络赛 ）

There are N light bulbs indexed from 0 to N−1. Initially, all of them are off.

A FLIP operation switches the state of a contiguous subset of bulbs. FLIP(L,R)means to flip all bulbs xsuch that L≤x≤R. So for example, FLIP(3,5) means to flip bulbs 3 , 4 and 5, and FLIP(5,5) means to flip bulb 5.

Given the value of N and a sequence of M flips, count the number of light bulbs that will be on at the end state.

InputFile

The first line of the input gives the number of test cases T. T test cases follow. Each test case starts with a line containing two integers N and M, the number of light bulbs and the number of operations, respectively. Then, there are M more lines, the i-th of which contains the two integers Li and Rii​, indicating that the i-th operation would like to flip all the bulbs from Li to Ri, inclusive.

1≤T≤1000

1≤N≤1e6

1≤M≤1000

0≤Li≤Ri≤N−1

OutputFile

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the number of light bulbs that will be on at the end state, as described above.

样例输入复制

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

样例输出复制

Case #1: 4
Case #2: 3

 

 题意：T组样例，给定N个灯泡，M个操作，[L,R]代表打开区间内灯泡，若灯泡为开状态就关闭。

            线段树超内存，这几天没课了总结下这个内存这个问题，真的坑人

思路： 若M=2,操作区间分别是[1, 2] [3, 4]  ans = (2+1)-1 + (4+1)-3  = 4 

            若M=2,操作区间分别是[1, 4] [2, 3]  ans = 2-1 + (4+1)-(3+1)  = 4  观察得，将右端点+1，操作的区间的左右端点排序得到的序列，只需要遍历一次m个操作，求前缀和

             一个区间若被操作偶数次相当于没有操作，只有操作奇数次的区间的灯的状态才改变，可以通过M次操作的左右端点判断有哪些区间被操作了奇数次。

 

//c++11 AC   c++14超时 
#include 
#include 
using namespace std;
const int maxn = 1e5+10;
int a[maxn];
int main() {
    int T;
    cin >> T;
    for(int k=1;k<=T;k++) {
	    int n,m;
	    int ans = 0;
	    cin >> n >> m;
	    for(int i=0;i<2*m;i++) {
	    	if(i%2==0)
				cin >> a[i];
			else {
				cin >> a[i];
				a[i]++;
			}
		}
	    	
		sort(a,a+2*m);
    	for(int i=0;i<2*m;i+=2) {
    		ans = ans+(a[i+1]-a[i]);
		}
    	
    	cout << "Case #" << k << ": " << ans << endl;
	}
    
    return 0;
}