压缩状态思想很不错的
每次把原来的贡献减掉，新来的再加上
最多涉及10个点
注意：
1.去重
2.下标从0开始
3.线段树初始的最小值个数都是r-l+1
代码：
#include<bits/stdc++.h>
#define reg register int
#define il inline
#define mid ((l+r)>>1)
#define numb (ch^'0')
using namespace std;
typedef long long ll;
il void rd(int &x){
char ch;x=0;bool fl=false;
while(!isdigit(ch=getchar()))(ch=='-')&&(fl=true);
for(x=numb;isdigit(ch=getchar());x=x*10+numb);
(fl==true)&&(x=-x);
}
namespace Miracle{
const int N=1000000+5;
const int inf=0x3f3f3f3f;
int id[N],vis[N],sta[N],top;
int pos[N][2];
int n,m,q;
int num(int x,int y){
return (x-1)*m+y;
}
struct tr{
int mi,cnt;
int mx;
int ad;
}t[4*N];
void pushup(int x){
t[x].mi=min(t[x<<1].mi,t[x<<1|1].mi);
t[x].mx=max(t[x<<1].mx,t[x<<1|1].mx);
t[x].cnt=t[x<<1].cnt*(t[x<<1].mi==t[x].mi)+t[x<<1|1].cnt*(t[x<<1|1].mi==t[x].mi);
}
void pushdown(int x){
if(!t[x].ad) return;
t[x<<1].ad+=t[x].ad;
t[x<<1].mi+=t[x].ad;
t[x<<1].mx+=t[x].ad;
t[x<<1|1].ad+=t[x].ad;
t[x<<1|1].mi+=t[x].ad;
t[x<<1|1].mx+=t[x].ad;
t[x].ad=0;
}
void build(int x,int l,int r){
if(l==r){
t[x].cnt=1;return;
}
build(x<<1,l,mid);
build(x<<1|1,mid+1,r);
pushup(x);
}
void add(int x,int l,int r,int L,int R,int c){
if(L<=l&&r<=R){
t[x].ad+=c;
t[x].mi+=c;
t[x].mx+=c;
if(t[x].mi==t[x].mx) t[x].cnt=r-l+1;
return;
}
pushdown(x);
if(L<=mid) add(x<<1,l,mid,L,R,c);
if(mid<R) add(x<<1|1,mid+1,r,L,R,c);
pushup(x);
}
int mv[4][2]={{+1,0},{-1,0},{0,+1},{0,-1}};
void wrk1(int x,int y,int c){
//cout<<" x y "<<x<<" "<<y<<" c "<<c<<endl;
int mi=inf,cmi=inf;
for(reg i=0;i<4;++i){
int dx=x+mv[i][0],dy=y+mv[i][1];
if(dx>=1&&dx<=n&&dy>=1&&dy<=m){
if(id[num(dx,dy)]<mi){
cmi=mi;mi=id[num(dx,dy)];
}else cmi=min(cmi,id[num(dx,dy)]);
}
}
if(cmi<=id[num(x,y)]-1) {
//cout<<" x y "<<x<<" "<<y<<" : "<<mi<<" "<<cmi<<endl;
add(1,1,n*m,cmi,id[num(x,y)]-1,c);
}
}
void wrk2(int x,int y,int c){
int mi=n*m+1;
int dx=x-1,dy=y;
if(dx>=1&&dx<=n&&dy>=1&&dy<=m){
mi=min(mi,id[num(dx,dy)]);
}
dx=x,dy=y-1;
if(dx>=1&&dx<=n&&dy>=1&&dy<=m){
mi=min(mi,id[num(dx,dy)]);
}
//cout<<" x y "<<x<<" "<<y<<" mimimi "<<mi<<endl;
if(id[num(x,y)]<=mi-1) add(1,1,n*m,id[num(x,y)],mi-1,c);
}
void sol(int x,int y,int c){
//cout<<" sol "<<x<<" "<<y<<" c "<<c<<endl;
if(!vis[num(x,y)]) {
sta[++top]=num(x,y);
vis[num(x,y)]=1;
wrk1(x,y,c);wrk2(x,y,c);
}
for(reg i=0;i<4;++i){
int dx=x+mv[i][0],dy=y+mv[i][1];
if(dx>=1&&dx<=n&&dy>=1&&dy<=m&&!vis[num(dx,dy)]){
vis[num(dx,dy)]=1;
sta[++top]=num(dx,dy);
wrk1(dx,dy,c);wrk2(dx,dy,c);
}
}
}
int main(){
rd(n);rd(m);rd(q);
build(1,1,n*m);
int x,y;
for(reg i=1;i<=n*m;++i){
rd(x);rd(y);
++x;++y;
pos[i][0]=x;pos[i][1]=y;
id[num(x,y)]=i;
}
for(reg i=1;i<=n;++i){
for(reg j=1;j<=m;++j){
wrk1(i,j,1);
wrk2(i,j,1);
}
}
//cout<<t[1].mi<<" "<<t[1].mx<<" "<<t[1].cnt<<endl;
int A,B;
while(q--){
rd(A);rd(B);
++A;++B;
top=0;
x=pos[A][0],y=pos[A][1];
sol(x,y,-1);
x=pos[B][0],y=pos[B][1];
sol(x,y,-1);
while(top) vis[sta[top--]]=0;
swap(id[num(pos[A][0],pos[A][1])],id[num(pos[B][0],pos[B][1])]);
swap(pos[A][0],pos[B][0]);
swap(pos[A][1],pos[B][1]);
x=pos[A][0],y=pos[A][1];
sol(x,y,1);
x=pos[B][0],y=pos[B][1];
sol(x,y,1);
while(top) vis[sta[top--]]=0;
int ans=0;
if(t[1].mi==1) ans=t[1].cnt;
printf("%d\n",ans);
}
return 0;
}
}
signed main(){
Miracle::main();
return 0;
}
/*
Author: *Miracle*
Date: 2019/2/10 17:53:31
*/
来源：https://www.cnblogs.com/Miracevin/p/10360146.html