HDU 5372 Segment Game

线段包含问题本身的解答是CDQ分治，时间复杂度为O(nlog(n)2)
但是这题会卡你的CDQ
但是因为这题它加的线段的长度是递增的，所以能够保证的是，之前加进去的线段，如果能够被包含的一定是完全被包含，同时也就满足了：右端点左边的右端点数 - 左端点-1左边的左端点数 = 完全包含的线段个数。
CDQ的时候这维需要人工去维护，但是因为这里已经给你排序好了，就可以直接乱搞了。

//      whn6325689//      Mr.Phoebe//      http://blog.csdn.net/u013007900#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define eps 1e-9#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LLINF 1LL<<62#define speed std::ios::sync_with_stdio(false);typedef long long ll;typedef unsigned long long ull;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define CPY(x,y) memcpy(x,y,sizeof(x))#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define ls (idx<<1)#define rs (idx<<1|1)#define lson ls,l,mid#define rson rs,mid+1,r#define root 1,1,ntemplate<class T>inline bool read(T &n){    T x = 0, tmp = 1;    char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------const int MAXN=200010;int n;int c1[MAXN<<1],c2[MAXN<<1];struct Node{    int op;    int l,r; }a[MAXN];void update(int i,int v,int *c){    for(;i<MAXN*2;i+=lowbit(i))        c[i]+=v;}int getsum(int i,int *c){    int sum=0;    for(;i;i-=lowbit(i))        sum+=c[i];    return sum;}int b[MAXN<<1],cnt,num;int pos[MAXN],pos2[MAXN];int main(){    //freopen("data.txt","r",stdin);    int cas=1;    while(read(n))    {        CLR(c1,0);CLR(c2,0);        int op,l,r;        cnt=0,num=0;        for(int i=0;i<n;i++)        {            read(op),read(l);            if(op==0)            {                num++;                b[cnt++]=l;                b[cnt++]=l+num;            }            a[i].op=op;a[i].l=l;a[i].r=l+num;        }        sort(b,b+cnt);        cnt=unique(b,b+cnt)-b;        num=1;        printf("Case #%d:\n",cas++);        for(int i=0;i<n;i++)        {            if(a[i].op==0)            {                l=lower_bound(b,b+cnt,a[i].l)-b+1;                r=lower_bound(b,b+cnt,a[i].r)-b+1;                pos[num]=l;pos2[num]=r;num++;                //cout<<" "<<l<<" "<<r<<endl;                int num1=getsum(l-1,c1);                int num2=getsum(r,c2);                write(num2-num1),putchar('\n');                update(l,1,c1);                update(r,1,c2);            }            else            {                l=a[i].l;                //cout<<pos[l]<<" "<<pos2[l]<<endl;                update(pos[l],-1,c1);                update(pos2[l],-1,c2);            }        }    }    return 0;}