B. Light bulbs（2019 ICPC上海站）

Posted 2022-12-02 lusiqi

There are NN light bulbs indexed from 00 to N-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)means to flip all bulbs xx such that L \leq x \leq RL≤x≤R. So for example, FLIP(3, 5)FLIP(3,5) means to flip bulbs 33 , 44 and 55, and FLIP(5, 5)FLIP(5,5) means to flip bulb 55.

Given the value of NN and a sequence of MMflips, 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, TT. TT test cases follow. Each test case starts with a line containing two integers NN and MM, the number of light bulbs and the number of operations, respectively. Then, there are MMmore lines, the ii-th of which contains the two integers L_iLi? and R_iRi?, indicating that the ii-th operation would like to flip all the bulbs from L_iLi? to R_iRi? , inclusive.

1 \leq T \leq 10001≤T≤1000

1 \leq N \leq 10^61≤N≤106

1 \leq M \leq 10001≤M≤1000

0 \leq L_i \leq R_i \leq N-10≤Li?≤Ri?≤N−1

OutputFile

For each test case, output one line containing Case #x: y, where xx is the test case number (starting from 11) and yy 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


代码：
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+20;
int num[maxn*2];
int main()
    int t;
    cin>>t;
    for(int i=1;i<=t;i++)
        int n,m;
        scanf("%d%d",&n,&m);
        int count = 0;
        for(int j = 0; j<m;j++)
            int x,y;
            scanf("%d%d",&x,&y);
            num[count++] = x;
            num[count++] = y+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",i,sum);    
    
    return 0;

思路：
有一点差分的思想，但是却进行了一定的改变