Crayon 线段树或者树状数组

Problem Description

Background

Mary love painting so much, but as we know she can't draw very well. There is no one appreciate her works, so she come up with a puzzle with herself.

Description

There are only one case in each input file, the first line is a integer N (N ≤ 1,000,00) denoted the total operations executed by Mary.

Then following N lines, each line is one of the folling operations.

D L R : draw a segment [L, R], 1 ≤ L ≤ R ≤ 1,000,000,000.
C I : clear the ith added segment. It’s guaranteed that the every added segment will be cleared only once.
Q L R : query the number of segment sharing at least a common point with interval [L, R]. 1 ≤ L ≤ R ≤ 1,000,000,000.

Input

Then following n operations ...

Output

For each query, print the result on a single line ...

Sample Input

Sample Output

2
2

线段树

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <math.h>
  4 #include <string.h>
  5 #include <algorithm>
  6 using namespace std;
  7 #define ll long long
  8 typedef struct abcd
  9 {
 10     int l,r,ci,i;
 11     char x;
 12 } abcd;
 13 abcd a[110000];
 14 typedef struct abc
 15 {
 16     int x,i;
 17 } abc;
 18 abc c[300000];
 19 int cn=0,b[110000],bn=1;
 20 bool cmp(abc x,abc y)
 21 {
 22     return x.x<y.x;
 23 }
 24 bool cmp1(abcd x,abcd y)
 25 {
 26     return x.l<y.l;
 27 }
 28 bool cmp2(abcd x,abcd y)
 29 {
 30     return x.r<y.r;
 31 }
 32 bool cmp3(abcd x,abcd y)
 33 {
 34     return x.i<y.i;
 35 }
 36 typedef struct tree
 37 {
 38     int a,d,sub;
 39 } tree;
 40 tree t[600000];
 41 void fun(int x)
 42 {
 43     if(t[x].d)
 44     {
 45         t[x<<1].d+=t[x].d;
 46         t[x<<1].sub+=t[x].d;
 47         t[x<<1|1].sub+=t[x].d;
 48         t[x<<1|1].d+=t[x].d;
 49         t[x<<1].a+=t[x].d;
 50         t[x<<1|1].a+=t[x].d;
 51         t[x].d=0;
 52     }
 53 }
 54 void update(int x,int y,int b,int c,int tt,int z)
 55 {
 56     if(x<=b&&y>=c)
 57     {
 58         t[tt].sub+=z;
 59         t[tt].d+=z;
 60         t[tt].a+=z;
 61         return ;
 62     }
 63     if(t[tt].d)
 64     fun(tt);
 65     int m=(b+c)>>1;
 66     if(x<=m&&y>m)t[tt].sub+=z;
 67     if(x<=m)update(x,y,b,m,tt<<1,z);
 68     if(y>m)update(x,y,m+1,c,tt<<1|1,z);
 69     t[tt].a=t[tt<<1].a+t[tt<<1|1].a-t[tt].sub;
 70 }
 71 int query(int x,int y,int b,int c,int tt)
 72 {
 73     if(x<=b&&y>=c)
 74     {
 75         return t[tt].a;
 76     }
 77     if(t[tt].d)
 78     fun(tt);
 79     int m=(b+c)>>1;
 80     int r=0,sub;
 81     if(x<=m)r=query(x,y,b,m,tt<<1);
 82     if(y>m)r=r+query(x,y,m+1,c,tt<<1|1);
 83     t[tt].a=t[tt<<1].a+t[tt<<1|1].a-t[tt].sub;
 84     if(x<=m&&y>m)
 85         return r-t[tt].sub;
 86     else return r;
 87 }
 88 int main()
 89 {
 90     int n,i,j;
 91    //freopen("in.txt","r",stdin);
 92     scanf("%d",&n);
 93     for(i=0; i<n; i++)
 94     {
 95         getchar();
 96         scanf("%c",&a[i].x);
 97         if(a[i].x=='C')
 98         {
 99             scanf("%d",&a[i].ci);
100         }
101         else
102         {
103             scanf("%d%d",&a[i].l,&a[i].r);
104             c[cn++].x=a[i].l,c[cn++].x=a[i].r;
105             if(a[i].x=='D')
106                 b[bn++]=i;
107         }
108         a[i].i=i;
109     }
110     int now=2;
111     sort(c,c+cn,cmp);
112     c[0].i=1;
113     for(i=1; i<cn; i++)
114     {
115         if(c[i].x==c[i-1].x)
116             c[i].i=c[i-1].i;
117         else c[i].i=now++;
118     }
119 
120     sort(a,a+n,cmp1);
121     j=0;
122     for(i=0; i<n; i++)
123     {
124         while(i<n&&a[i].x=='C')i++;
125         if(i==n)break;
126         while(a[i].l!=c[j].x)j++;
127         a[i].l=c[j].i;
128     }
129     sort(a,a+n,cmp2);
130     j=0;
131     for(i=0; i<n; i++)
132     {
133         while(i<n&&a[i].x=='C')i++;
134         if(i==n)break;
135         while(a[i].r!=c[j].x)j++;
136         a[i].r=c[j].i;
137     }
138     sort(a,a+n,cmp3);
139     /*for(i=0; i<n; i++)
140         cout<<a[i].x<<" "<<a[i].l<<" "<<a[i].r<<endl;*/
141     memset(t,0,sizeof(t));
142     for(i=0; i<n; i++)
143     {
144         if(a[i].x=='D')
145         {
146             update(a[i].l,a[i].r,1,c[cn-1].i,1,1);
147         }
148         else if(a[i].x=='C')
149         {
150             update(a[b[a[i].ci]].l,a[b[a[i].ci]].r,1,c[cn-1].i,1,-1);
151         }
152         else
153         {
154             printf("%d
",query(a[i].l,a[i].r,1,c[cn-1].i,1));
155         }
156     }
157 }

View Code

树状数组

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <math.h>
  4 #include <string.h>
  5 #include <algorithm>
  6 using namespace std;
  7 #define ll long long
  8 typedef struct abcd
  9 {
 10     int l,r,ci,i;
 11     char x;
 12 } abcd;
 13 abcd a[110000];
 14 typedef struct abc
 15 {
 16     int x,i;
 17 } abc;
 18 abc c[300000];
 19 int cn=0,b[110000],bn=1;
 20 bool cmp(abc x,abc y)
 21 {
 22     return x.x<y.x;
 23 }
 24 bool cmp1(abcd x,abcd y)
 25 {
 26     return x.l<y.l;
 27 }
 28 bool cmp2(abcd x,abcd y)
 29 {
 30     return x.r<y.r;
 31 }
 32 bool cmp3(abcd x,abcd y)
 33 {
 34     return x.i<y.i;
 35 }
 36 int ab[800000][2],m;
 37 int lowbit(int x)
 38 {
 39     return x&(-x);
 40 }
 41 void update(int y,int x,int z)
 42 {
 43     while(x<=m)
 44     {
 45         ab[x][y]+=z;
 46         x+=lowbit(x);
 47     }
 48 }
 49 int query(int y,int x)
 50 {
 51     int sum=0;
 52     while(x>0)
 53     {
 54         sum+=ab[x][y];
 55         x-=lowbit(x);
 56     }
 57     return sum;
 58 }
 59 int main()
 60 {
 61     int n,i,j;
 62    // freopen("in.txt","r",stdin);
 63     scanf("%d",&n);
 64     for(i=0; i<n; i++)
 65     {
 66         getchar();
 67         scanf("%c",&a[i].x);
 68         if(a[i].x=='C')
 69         {
 70             scanf("%d",&a[i].ci);
 71         }
 72         else
 73         {
 74             scanf("%d%d",&a[i].l,&a[i].r);
 75             c[cn++].x=a[i].l,c[cn++].x=a[i].r;
 76             if(a[i].x=='D')
 77                 b[bn++]=i;
 78         }
 79         a[i].i=i;
 80     }
 81     int now=2,sum=0;
 82     sort(c,c+cn,cmp);
 83     c[0].i=1;
 84     for(i=1; i<cn; i++)
 85     {
 86         if(c[i].x==c[i-1].x)
 87             c[i].i=c[i-1].i;
 88         else c[i].i=now++;
 89     }
 90 
 91     sort(a,a+n,cmp1);
 92     j=0;
 93     for(i=0; i<n; i++)
 94     {
 95         while(i<n&&a[i].x=='C')i++;
 96         if(i==n)break;
 97         while(a[i].l!=c[j].x)j++;
 98         a[i].l=c[j].i;
 99     }
100     sort(a,a+n,cmp2);
101     j=0;
102     for(i=0; i<n; i++)
103     {
104         while(i<n&&a[i].x=='C')i++;
105         if(i==n)break;
106         while(a[i].r!=c[j].x)j++;
107         a[i].r=c[j].i;
108     }
109     sort(a,a+n,cmp3);
110     /*for(i=0; i<n; i++)
111         cout<<a[i].x<<" "<<a[i].l<<" "<<a[i].r<<endl;*/
112     m=c[cn-1].i;
113     for(i=0; i<n; i++)
114     {
115         if(a[i].x=='D')
116         {
117             update(0,a[i].l,1);
118             update(1,a[i].r,1);
119             sum++;
120         }
121         else if(a[i].x=='C')
122         {
123             update(0,a[b[a[i].ci]].l,-1);
124             update(1,a[b[a[i].ci]].r,-1);
125             sum--;
126         }
127         else
128         {
129             int ans=query(0,a[i].r);
130             ans-=query(1,a[i].l-1);
131             printf("%d
",ans);
132         }
133     }
134 }

View Code