HDU 4879 ZCC loves march

On a m*m land stationed n troops, numbered from 1 to n. The i-th troop's position can be described by two numbers (xi,yi) (1<=xi<=m,1<=yi<=m). It is possible that more than one troop stationed in the same place. 

 Then there are t minutes, in each minute one of the following two events will occur:

 (1)the x-th troop moves towards a direction( Up(U) Down(D) Left(L) Right(R))for d units;(You can suppose that the troops won't move out of the boundary) 

 (2)the x-th troop needs to regroup the troops which stations in the same row or column with the x-th troop. That is, these troops need to move to the x-th troop's station. 

 Suggest the cost of i-th troop moving to the j-th troop is (xi-xj)^2+(yi-yj)^2, every time a troop regroups, you should output the cost of the regrouping modulo 10^9+7. 

InputThe first line: two numbers n,m(n<=100000,m<=10^18) 

Next n lines each line contain two numbers xi,yi(1<=xi,yi<=m) 

Next line contains a number t.(t<=100000) 

Next t lines, each line's format is one of the following two formats: 

(1)S x d, S∈{U,L,D,R}, indicating the first event(1<=x<=n,0<=d<m) 

(2)Q x, indicating the second event(1<=x<=n) 

 In order to force you to answer the questions online, each time you read x', x=x'�lastans("�" means "xor"), where lastans is the previous answer you output. At the beginning lastans=0. 

OutputQ lines, i-th line contain your answer to the i-th regrouping event.(modulo 10^9+7)
Sample Input

5 3
1 3
2 1
2 2
2 3
3 2
6
Q 1
L 0 2
L 5 2
Q 5
R 3 1
Q 3

Sample Output

1
1
7

Hint

The input after decode:
Q 1
L 1 2
L 4 2
Q 4
R 2 1
Q 2



将每次的移动，看做是一个新的点的诞生，合并使用并查集，用两个map套个容器就可以了。#include<map>
#include<set>
#include<ctime>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define ms(x,y) memset(x,y,sizeof(x))
#define rep(i,j,k) for(int i=j;i<=k;i++)
#define per(i,j,k) for(int i=j;i>=k;i--)
#define loop(i,j,k) for (int i=j;i!=-1;i=k[i])
#define inone(x) scanf("%d",&x)
#define intwo(x,y) scanf("%d%d",&x,&y)
#define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p)
#define lson x<<1,l,mid
#define rson x<<1|1,mid+1,r
#define mp(i,j) make_pair(i,j)
#define ft first
#define sd second
typedef long long LL;
typedef pair<LL, LL> pii;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;
const double eps = 1e-10;
int T, n, m, fa
, cnt
, sz, g
, o;
LL d;
pii a
;
map<LL, set<int>> x, y;
char s
;

int get(int x) { return x == fa[x] ? x : fa[x] = get(fa[x]); }

int main()
{
while (~scanf("%d%lld", &n, &d))
{
for (auto i : x) i.second.clear();
for (auto i : y) i.second.clear();
x.clear(); y.clear();
rep(i, 1, n)
{
cnt[i] = 1; fa[i] = g[i] = i;
scanf("%lld%lld", &a[i].ft, &a[i].sd);
x[a[i].ft].insert(i); y[a[i].sd].insert(i);
}
sz = n;
inone(m);
int last = 0, ans;
while (m--)
{
scanf("%s", s);
inone(o); o = o^last;
int fo = get(g[o]);
if (s[0] == 'Q')
{
ans = 0;
for (auto i : x[a[fo].ft])
{
if (i == fo) continue;
LL q = (a[i].sd - a[fo].sd) % mod;
ans = (q * q % mod * cnt[i] + ans) % mod;
cnt[fo] += cnt[i]; fa[i] = fo; cnt[i] = 0;
}
for (auto i : y[a[fo].sd])
{
if (i == fo) continue;
LL q = (a[i].ft - a[fo].ft) % mod;
ans = (q * q % mod * cnt[i] + ans) % mod;
cnt[fo] += cnt[i]; fa[i] = fo; cnt[i] = 0;
}
x[a[fo].ft].clear(); x[a[fo].ft].insert(fo);
y[a[fo].sd].clear(); y[a[fo].sd].insert(fo);
printf("%d\n", last = ans); continue;
}
scanf("%lld", &d);
++sz;
cnt[sz] = 1; fa[sz] = sz; cnt[fo]--;
a[sz] = a[fo]; g[o] = sz;
if (s[0] == 'L' || s[0] == 'R') a[sz].sd += s[0] == 'L' ? -d : d;
else a[sz].ft += s[0] == 'U' ? -d : d;
x[a[sz].ft].insert(sz);
y[a[sz].sd].insert(sz);
}
}
return 0;
}