题目链接：http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2167
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
思路：在纸上画一下图，可以发现答案跟a,b所跨水平和垂直区间有关，如图，假设nn为水平方向所占用的区间，mm为垂直方向所占的区间，且共有n行m列，则它是把mm×m+nn×n-nn×mm个点连通成一个点，所以答案ans=n×m-(mm×m+nn×n-nn×mm)+1=n×m-mm×m-nn×n+nn×mm+1,但注意当水平方向占的区间为0时，是将mm行的点连通成mm个点，即mm×m个点连通成mm个点,此时ans=n×m-mm×m+mm,当垂直方向区间为0时也可想而知；
AC代码：
#include<bits/stdc++.h>
using namespace std;
struct node
{
int x,y;
node*l,*r;
}*t1,*t2;
int Insert(int x,int y,node *t)
{
if(t->x==0){t->x=x;t->y=y;return y-x+1;}
if(x>=t->x&&y<=t->y) return 0;
if(y<t->x)
{
if(t->l==NULL)
{
node *s=(node*)malloc(sizeof(node));
s->l=s->r=NULL;
s->x=0;
t->l=s;
}
return Insert(x,y,t->l);
}
if(x>t->y)
{
if(t->r==NULL)
{
node *s=(node*)malloc(sizeof(node));
s->l=s->r=NULL;
s->x=0;
t->r=s;
}
return Insert(x,y,t->r);
}
int ans=0;
if(x<t->x)
{
if(t->l==NULL)
{
node *s=(node*)malloc(sizeof(node));
s->l=s->r=NULL;
s->x=0;
t->l=s;
}
ans+=Insert(x,t->x-1,t->l);
}
if(y>t->y)
{
if(t->r==NULL)
{
node *s=(node*)malloc(sizeof(node));
s->l=s->r=NULL;
s->x=0;
t->r=s;
}
ans+=Insert(t->y+1,y,t->r);
}
return ans;
}
int main()
{
int n,m,q;
while(scanf("%d%d%d",&n,&m,&q)!=EOF)
{
long long ans=(long long)n*m;
int opt,x,y;
long long nn=0,mm=0;
t1=(node*)malloc(sizeof(node));
t1->x=0;t1->l=t1->r=NULL;
t2=(node*)malloc(sizeof(node));
t2->x=0;t2->l=t2->r=NULL;
while(q--)
{
int opt,x,y;
scanf("%d%d%d",&opt,&x,&y);
if(opt==1)
{
mm+=Insert(x,y,t1);
if(nn==0)
printf("%lld\n",ans-mm*m+mm);
else printf("%lld\n",ans-mm*m-nn*n+nn*mm+1);
}
else
{
nn+=Insert(x,y,t2);
if(mm==0)
printf("%lld\n",ans-nn*n+nn);
else printf("%lld\n",ans-mm*m-nn*n+nn*mm+1);
}
}
}
}