There are N light bulbs indexed from 00 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≤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 Ri, indicating that the ii-th operation would like to flip all the bulbs from Li to Ri , inclusive.
1≤T≤1000
1 ≤N≤1000000
1 ≤M≤1000
0≤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 yyis 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
#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
const int maxn=2e3+5;
int str[maxn],str2[maxn],c[1000005],n,m;
int ask(int x)
{
int ans=0;
for(; x; x-=x&-x)ans+=c[x];
return ans;
}
void add(int x,int y)
{
for(; x<=n; x+=x&-x)c[x]+=y;
}
template<class T>
inline void read(T&x)
{
T ans=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
ans=ans*10+ch-'0';
ch=getchar();
}
x=ans*f;
}
int main()
{
int t,l,r,cnt=0,k=0;
read(t);
while(t--)
{
cnt=0;
read(n);
read(m);
int m2=m;
while(m2--)
{
read(l);read(r);
++l,++r;
add(l,1);
add(r+1,-1);
str[++cnt]=l,str2[cnt]=l;
str[++cnt]=r+1,str2[cnt]=r+1;
}
sort(str+1,str+cnt+1);
int cnt2=0;
for(int i=1; i<=cnt; ++i)
{
if(str[i]!=str[cnt2])
str[++cnt2]=str[i];
}
int ans=0,flag;
for(int i=1; i<cnt2; ++i)
{
flag=ask(str[i]);
if((flag&1)==1)
ans+=str[i+1]-1-str[i]+1;
}
printf("Case #%d: %d\n",++k,ans);
for(int i=1; i<=2*m; i+=2)
{
add(str2[i],-1);
add(str2[i+1],1);
}
}
return 0;
}