2014 Multi-University Training Contest 2 - 1008 (ZCC loves march)
http://acm.hdu.edu.cn/showproblem.php?pid=4879
ZCC loves march
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 130712/130712 K (Java/Others)
Total Submission(s): 762 Accepted Submission(s): 164
Problem 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.
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.
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
![](http://acm.hdu.edu.cn/data/images/C503-1008-1.png)
Hint
The input after decode:
Q 1
L 1 2
L 4 2
Q 4
R 2 1
Q 2![](http://acm.hdu.edu.cn/data/images/C503-1008-1.png)
Author
镇海中学
Source
题目大意: n个军团, 两种操作, 一种是编号为x的军团移动一步, 一个是编号为x的军团这行这列的所有军团都移动到x的位置, 并输出花费. 强制在线.
之前没有用并查集一直tle, 后来想想用并查集时间优化还是相当可观的. 也就是第二个操作将所有移动到x的棋子的根指向x, map的那一行那一列只留x一个, 并用结构(我用的plli)存储这个点集位置和里面有多少军团. 用map来维护每行每列有哪些根, 第一种操作就正常地查找x所在的根(如果只有x一个元素就删除)移动插入; 第二种操先查找x所在的根获得, 之后遍历改行该列, 更新ret值,将所有点指向一个新的根(cnt++), 删掉map<int,set<int> >row,col中已经移走的节点(这里居然开始就没想好.. 遍历行的时候需要随时删掉对应列的点;反之亦然. 最后还要将这一行一列clear) , 然后插入唯一的新的根, 结束操作.
虽然是写之前看明白的单写出来还是跟大牛写的顺序一模一样...不看还是不会..http://blog.csdn.net/faithdmc/article/details/38223667
#include <iostream> #include <algorithm> #include <string> #include <vector> #include <set> #include <map> #include <cstring> using namespace std; #define rep(i, n) for (int i = 0, _n = (int)(n); i < _n; i++) #define fer(i, x, n) for (int i = (int)(x), _n = (int)(n); i < _n; i++) #define mkp make_pair #define X first #define Y second template<class T> inline void smin(T &a, T b){if(b<a)a=b;} template<class T> inline void smax(T &a, T b){if(a<b)a=b;} typedef long long ll; typedef pair<ll,ll> pll; const int mod = 1000000007; //////////////////////////////////////////////////////////////////////////////// typedef pair<pll,int> plli; ll n,lastans,x,y,T,tmp,cnt,m,d; plli troop[200009]; map<ll,set<int> > col,row; int ff[200009]; set<int>::iterator ite; char c; int findroot(int x){ if(ff[x]!=x) ff[x]=findroot(ff[x]); return ff[x]; } void move(){ int root = findroot(x); ll tx=troop[root].X.X,ty=troop[root].X.Y; troop[root].Y--; if(troop[root].Y==0) row[troop[root].X.X].erase(root),col[troop[root].X.Y].erase(root); switch(c){ case 'U': tx-=d; break; case 'D': tx+=d; break; case 'L': ty-=d; break; case 'R': ty+=d; break; } ff[x]=x; troop[x]=mkp(mkp(tx,ty),1); row[tx].insert(x); col[ty].insert(x); } inline ll f(ll x){ return (x>0?(x%mod):((-x)%mod));} ll query(){ ll ret=0; int root = findroot(x),num = 0; ll tx=troop[root].X.X,ty=troop[root].X.Y; for(ite = row[tx].begin();ite!=row[tx].end();ite++){ ret= f(ret+ f(troop[*ite].Y * f(f(troop[*ite].X.Y-ty)*f(troop[*ite].X.Y-ty)))),num+=troop[*ite].Y,ff[*ite]=cnt,col[troop[*ite].X.Y].erase(*ite); } for(ite = col[ty].begin();ite!=col[ty].end();ite++){ ret= f(ret+ f(troop[*ite].Y * f(f(troop[*ite].X.X-tx)*f(troop[*ite].X.X-tx)))),num+=troop[*ite].Y,ff[*ite]=cnt,row[troop[*ite].X.X].erase(*ite); } troop[cnt]=mkp(mkp(tx,ty),num); ff[cnt]=cnt; row[tx].clear(); col[ty].clear(); row[tx].insert(cnt); col[ty].insert(cnt); cnt++; return ret; } int main() { //freopen("in.txt","r",stdin); ios_base::sync_with_stdio(0); while(scanf("%I64d %I64d",&n,&m)!=EOF){ row.clear(); col.clear(); lastans=0; cnt = n+1; fer(i,1,n+1) scanf("%I64d %I64d",&troop[i].X.X,&troop[i].X.Y),ff[i]=i,row[troop[i].X.X].insert(i),col[troop[i].X.Y].insert(i),troop[i].Y=1; scanf("%I64d",&T); rep(tt,T){ scanf("\n%c %I64d",&c,&x); x^=lastans; if(c!='Q') scanf("%I64d",&d),move(); else lastans = query(),printf("%I64d\n",lastans); } } return 0; }