题目链接:
ACdream 1055
题意:
给你
题解:
先离散化处理一下,再用树状数组维护一下区间变化即可。
详细看代码。
AC代码:
#include <bits/stdc++.h>
using namespace std;
struct data
{
int x,y;
char ch;
}d[123456];
int tree[200010][2];
int has[200010],all;
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int val,int k)
{
while(x<=all)
{
tree[x][k] += val;
x+=lowbit(x);
}
}
int query(int x,int k)
{
int res = 0 ;
while(x)
{
res += tree[x][k];
x -= lowbit(x);
}
return res;
}
int solve(int l,int r)
{
return query(r,0) - query(l-1,1);
}
int main()
{
int n;
while(~scanf("%d",&n))
{
all = 0;
for(int i=1;i<=n;i++)
{
getchar();
scanf("%c",&d[i].ch);
if(d[i].ch=='C')
{
scanf("%d",&d[i].x);
has[all++] = d[i].x;
}
else // D , Q
{
scanf("%d%d",&d[i].x,&d[i].y);
has[all++] = d[i].x;
has[all++] = d[i].y;
}
}
sort(has,has+all);
all = unique(has,has+all) - has;
memset(tree,0,sizeof(tree));
int cnt = 0;
for(int i=1;i<=n;i++)
{
if(d[i].ch =='D')//draw
{
int l = lower_bound(has, has+all, d[i].x) - has;
int r = lower_bound(has, has+all, d[i].y) - has;
update(l+1,1,0);
update(r+1,1,1);
d[++cnt] = d[i];
}
else if(d[i].ch == 'Q') //query
{
int l = lower_bound(has,has+all,d[i].x) - has;
int r = lower_bound(has,has+all,d[i].y) - has;
printf("%d\n",solve(l+1,r+1));
}
else if(d[i].ch == 'C') //clear
{
int l = lower_bound(has,has+all,d[d[i].x].x) - has;
int r = lower_bound(has,has+all,d[d[i].x].y) - has;
update(l+1,-1,0);
update(r+1,-1,1);
}
}
}
return 0;
}