2019上海ICPC网络赛B Light bulbs（差分+优化）

时间限制：1000ms    内存：8192K

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 $x$ such that $L\le x\le R$. 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 $L^i$ and $R^i$, indicating that the $i$-th operation would like to flip all the bulbs from $L^i$ to $R^i$ , inclusive.

$1\le T\le 1000$

$1\le N\le 10^6$

$1\le M\le 1000$

$0\le L^i\le R^i\le N-1$

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

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

样例输出

Case #1: 4

Case #2: 3


题目大意：
n个灯，初始全为不亮，共m个操作，每次操作包含两个数l和r，表示改变该区间内所有灯的状态（亮->不亮，不亮->亮）

思路：
这题如果线段树去写的话会爆内存，差不多只能开两倍的内存
用到差分的思想（左端点位置的值+1，右端点+1位置的值-1），考虑到m比较小，n比较大，差分后直接求前缀和的话还是会超时，所以需要考虑对m进行操作
对每个修改的点按位置进行排序，用sum求和，当sum为奇数时，说明与前一个位置相差奇数次操作，相差的区间内灯全亮

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N=1e6+5,M=2005;
int a[N];

struct Node{
    int index,num;
}node[M];

bool cmp(Node x,Node y){
    return x.index<y.index;
}

int main(){
    int t,n,m,l,r,ca=0;
    scanf("%d",&t);
    while(t--){
        int k=0;
        memset(a,0,sizeof a);
        scanf("%d%d",&n,&m);
        while(m--){
            scanf("%d%d",&l,&r);
            node[k].index=l,node[k++].num=1;
            node[k].index=r+1,node[k++].num=-1;
        }
        sort(node,node+k,cmp);
        int sum=0,res=0;
        for(int i=1;i<k;i++){
            sum+=node[i].num;
            if(sum&1){
                res+=node[i].index-node[i-1].index;
            }
        }
        printf("Case #%d: %d\n",++ca,res);
    }
}