Description
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.
Input
The 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.
Output
Q 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
思路：分别用map来维护与x坐标平行和垂直的线上面的集合，集合里面直接存对应元素的下标，并用并查集维护，根节点可以用来表示所在的集合。当移动一个元素时，先找到该元素下标对应的根，根对应的num-1，并把移动的元素插到新的位置上，他的根就是他自己。执行Q操作时，通过map找到同一行和同一列的所有集合进行计算，再把这些计算过的集合删掉，同时下标指向新的根（即移动之后形成的新的集合）。
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#define LL long long
#define mod 1000000007
using namespace std;
struct TROOP{
LL x,y,num;
TROOP(){}
TROOP(LL nx,LL ny,LL nnum){x=nx,y=ny,num=nnum;}
}troop[200005];
map<LL,set<LL> >MX;
map<LL,set<LL> >MY;
set<LL>::iterator it;
LL node[200005];
LL findroot(LL x)
{
if(node[x]!=x) node[x]=findroot(node[x]);
return node[x];
}
int main()
{
LL m,x,y,n,i,t,cnt,root,a,b,ans;
char s[5];
while(~scanf("%I64d%I64d",&n,&m))
{
MX.clear();
MY.clear();
for(i=1;i<=n;i++)
{
scanf("%I64d%I64d",&x,&y);
troop[i].x=x;
troop[i].y=y;
troop[i].num=1;
node[i]=i;
MX[x].insert(i);
MY[y].insert(i);
}
cnt=n+1;
ans=0;
scanf("%I64d",&t);
while(t--)
{
scanf("%s",s);
if(s[0]=='Q')
{
scanf("%I64d",&a);
a^=ans;
root=findroot(a);//找到a所在的集合
x=troop[root].x;
y=troop[root].y;
LL num=0;
ans=0;
for(it=MX[x].begin();it!=MX[x].end();it++)
{
num+=troop[*it].num;
LL temp=abs(troop[*it].y-y);
temp%=mod;
ans=(temp*temp%mod*troop[*it].num%mod+ans)%mod;
node[*it]=cnt;//指向cnt，cnt是执行Q操作之后新的根，用来标记新的集合
MY[troop[*it].y].erase(*it);//*it已经计算过，从MY[]集合里删掉，避免重复计算
}
for(it=MY[y].begin();it!=MY[y].end();it++)
{
num+=troop[*it].num;
LL temp=abs(troop[*it].x-x);
temp%=mod;
ans=(temp*temp%mod*troop[*it].num%mod+ans)%mod;
node[*it]=cnt;//同理
MX[troop[*it].x].erase(*it);//同理
}
node[cnt]=cnt;//指向自己，别忘了
troop[cnt]=TROOP(x,y,num);//执行Q操作之后形成的新集合
MX[x].clear();
MY[y].clear();
MX[x].insert(cnt);//在目标集合中插入
MY[y].insert(cnt);
cnt++;
printf("%I64d\n",ans);
}
else
{
scanf("%I64d%I64d",&a,&b);
a^=ans;
root=findroot(a);//找到a所在的集合，即a的根节点
x=troop[root].x;
y=troop[root].y;
troop[root].num--;//集合里的计数减一
if(!troop[root].num)//如果集合的计数为0则把该集合删掉
{
MX[x].erase(root);
MY[y].erase(root);
}
if(s[0]=='U')
{
troop[a]=TROOP(x-b,y,1);
node[a]=a;//a指向自己，作为新的根
MX[x-b].insert(a);//在目标位置插入
MY[y].insert(a);
}
else if(s[0]=='L')//以下同理
{
troop[a]=TROOP(x,y-b,1);
node[a]=a;
MX[x].insert(a);
MY[y-b].insert(a);
}
else if(s[0]=='D')
{
troop[a]=TROOP(x+b,y,1);
node[a]=a;
MX[x+b].insert(a);
MY[y].insert(a);
}
else if(s[0]=='R')
{
troop[a]=TROOP(x,y+b,1);
node[a]=a;
MX[x].insert(a);
MY[y+b].insert(a);
}
}
}
}
}
HDU-4879-ZCC loves march(map+set+并查集)