题目链接
题解
线段树裸题，，对每一行每一列开线段树
由于坐标很大，用(map)维护根下标
化一下式子，只用维护区间和，区间平方和，区间存在的个数
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b,c) (node){a,b,c}
#define cls(s) memset(s,0,sizeof(s))
#define cp node
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 8000005,INF = 1000000000,P = 1000000007;
inline LL read(){
LL out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
struct node{LL a,b,c;};
map<LL,int> rt1,rt2;
LL x[maxn],y[maxn],m;
LL sum[maxm],sum2[maxm],num[maxm],cnt;
int ls[maxm],rs[maxm],n;
void upd(int u){
sum[u] = (sum[ls[u]] + sum[rs[u]]) % P;
sum2[u] = (sum2[ls[u]] + sum2[rs[u]]) % P;
num[u] = num[ls[u]] + num[rs[u]];
}
void modify(int& u,int l,int r,int pos,LL v,LL vv){
if (!u) u = ++cnt;
if (l == r){v %= P; sum[u] = v; sum2[u] = v * v % P; num[u] = vv; return;}
int mid = l + r >> 1;
if (mid >= pos) modify(ls[u],l,mid,pos,v,vv);
else modify(rs[u],mid + 1,r,pos,v,vv);
upd(u);
}
cp query(int u,int l,int r,int L,int R){
if (!u) return mp(0,0,0);
if (l >= L && r <= R) return mp(sum[u],sum2[u],num[u]);
int mid = l + r >> 1;
if (mid >= R) return query(ls[u],l,mid,L,R);
if (mid < L) return query(rs[u],mid + 1,r,L,R);
cp t1 = query(ls[u],l,mid,L,R),t2 = query(rs[u],mid + 1,r,L,R);
return mp((t1.a + t2.a) % P,(t1.b + t2.b) % P,t1.c + t2.c);
}
LL lans,t,l,r,d,X,Y;
int main(){
n = read(); m = read();
REP(i,n){
x[i] = read(); y[i] = read();
modify(rt1[x[i]],1,n,i,y[i] % P,1);
modify(rt2[y[i]],1,n,i,x[i] % P,1);
}
int T = read(); char opt; cp u;
while (T--){
opt = getchar(); while (!isalpha(opt)) opt = getchar();
if (opt == 'Q'){
t = read() ^ lans; l = read(); r = read();
u = query(rt1[x[t]],1,n,l,r);
X = x[t] % P; Y = y[t] % P;
lans = ((Y * Y % P * u.c % P - 2ll * Y % P * u.a % P) % P + u.b)% P;
u = query(rt2[y[t]],1,n,l,r);
lans = (lans + (X * X % P * u.c % P - 2ll * X % P * u.a % P) % P + u.b) % P;
lans = (lans % P + P) % P;
printf("%lld
",lans);
}
else {
t = read() ^ lans; d = read();
modify(rt1[x[t]],1,n,t,0,0);
modify(rt2[y[t]],1,n,t,0,0);
switch(opt){
case 'U':y[t] += d;break;
case 'D':y[t] -= d;break;
case 'L':x[t] -= d;break;
case 'R':x[t] += d;break;
default:break;
}
modify(rt1[x[t]],1,n,t,y[t] % P,1);
modify(rt2[y[t]],1,n,t,x[t] % P,1);
}
}
return 0;
}