标签：int s2 s1 ans 1ll Grid 2018 ti 2167
2167: Grid
Submit Page Summary Time Limit: 2 Sec Memory Limit: 128 Mb Submitted: 203 Solved: 32
Description
Bobo has n × m points arranged into a matrix with n rows and m columns. The points in the intersection of the i-th row and the j-th column is labeled with (i, j). He is going to perform q operations of the following 2 kinds.
- Given parameters a, b, add edges between points (i, j) and (i, j + 1) for all a ≤ i ≤ b and 1 ≤ j < m.
- Given parameters a, b, add edges between points (i, j) and (i + 1, j) for all 1 ≤ i < n and a ≤ j ≤ b.
Bobo would like to know the number of connected components after each operation.
- 1 ≤ n, m ≤ 109
- 1 ≤ q ≤ 105
- ti ∈ {1, 2}
- If ti = 1, 1 ≤ ai ≤ bi ≤ n. If ti = 2, 1 ≤ ai ≤ bi ≤ m.
- The sum of q does not exceed 5 × 105.
Input
The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains three integers n, m and q.
The i-th of the following q lines contains three integers ti, ai and bi where ti is the kind of the operation i and ai, bi are corresponding parameters.
Output
For each test case, print q integers which denote the number of connected components after each operation.
Sample Input
2 3 3 2 1 1 2 3 3 1 1 1 1000000000 1000000000 1 1 1 2
Sample Output
5 4 2 999999998000000002
Hint
Source
2018湖南省第14届大学生计算机程序设计竞赛
Author
ftiasch
今天学习了如何动态开点的线段树，感觉好妙啊，普通的线段树会由于数组大小的限制到达不了1e9，而这里正好需要从1到1e9且不能离散化。于是采用了动态开点，类似于主席树方法给每个节点赋值。只要更新的节点不会超过很多就可以了。
普通的线段树是这样的：
而动态开点后：
贴AC代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+10;
int s1[N],ls[N],rs[N],cnt1,rt1;
int s2[N],ls2[N],rs2[N],cnt2,rt2;
int n,m,q;
void init()
{
for(int i=0;i<=cnt1;i++)
s1[i]=ls[i]=rs[i]=0;
for(int i=0;i<=cnt2;i++)
s2[i]=ls2[i]=rs2[i]=0;
cnt1=cnt2=rt1=rt2=0;
}
void up1(int &o,int l,int r,int ql,int qr)
{
if(!o) o=++cnt1;
if(s1[o]==r-l+1) return ;
if(ql<=l&&r<=qr)
{
s1[o]=r-l+1;
return ;
}
int mid=l+r>>1;
if(ql<=mid) up1(ls[o],l,mid,ql,qr);
if(qr>mid) up1(rs[o],mid+1,r,ql,qr);
s1[o]=s1[ls[o]]+s1[rs[o]];
}
void up2(int &o,int l,int r,int ql,int qr)
{
if(!o) o=++cnt2;
if(s2[o]==r-l+1) return ;
if(ql<=l&&r<=qr)
{
s2[o]=r-l+1;
return ;
}
int mid=r+l>>1;
if(ql<=mid) up2(ls2[o],l,mid,ql,qr);
if(qr>mid) up2(rs2[o],mid+1,r,ql,qr);
s2[o]=s2[ls2[o]]+s2[rs2[o]];
}
int main()
{
while(~scanf("%d%d%d",&n,&m,&q))
{
init();
while(q--)
{
int op,l,r;
scanf("%d%d%d",&op,&l,&r);
if(op==1)up1(rt1,1,n,l,r);
else up2(rt2,1,m,l,r);
ll ans;
if(s1[1]==0) ans=1ll*(m-s2[1])*n+s2[1];
if(s2[1]==0) ans=1ll*(n-s1[1])*m+s1[1];
if(s1[1]!=0&&s2[1]!=0)
{
ans=1ll*n*m-(1ll*s1[1]*m+1ll*s2[1]*n-1ll*s1[1]*s2[1])+1;
}
printf("%lld\n",ans);
}
}
}
标签：int,s2,s1,ans,1ll,Grid,2018,ti,2167 来源： https://blog.csdn.net/qq_41286356/article/details/89094259
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。