HDU-4879-ZCC loves march(map+set+并查集)

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 31 32 12 22 33 26Q 1L 0 2L 5 2Q 5R 3 1Q 3 

 

Sample Output

 117 

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 1000000007using 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);                }            }        }    }}