HDU4631+Set+最近点对

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题意：一个空平面，每次增加一个点，
其坐标根据上一个点算出：(x[i-1] * Ax + Bx ) mod Cx，(y[i-1] * Ay + By ) mod Cy

求出现有点集中的最近点对的距离的平方，共增加n个点，求每次求得的平方的和

http://blog.csdn.net/liuledidai/article/details/9664031

/*每次对于一个新插入的点，找出x与之最近的，然后分别向两侧搜*/#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>#include<iostream>#include<queue>#include<map>#include<set>#include<math.h>using namespace std;typedef long long int64;//typedef __int64 int64;const int maxn = 5e5+5;const int64 inf = 999999999999;const double pi=acos(-1.0);const double eps = 1e-8;struct Node{int64 x,y;};Node cur,nxt;typedef pair<int64,int64> PII;#define MP(a,b) make_pair((a),(b)) set< pair<int64,int64> > s;int main(){int T;scanf("%d",&T);while( T-- ){int64 Ax,Bx,Cx,Ay,By,Cy;int n;scanf("%d%I64d%I64d%I64d%I64d%I64d%I64d",&n,&Ax,&Bx,&Cx,&Ay,&By,&Cy);//scanf("%d%lld%lld%lld%lld%lld%lld",&n,&Ax,&Bx,&Cx,&Ay,&By,&Cy);s.clear();cur.x = cur.y = 0;nxt.x = (Ax*cur.x+Bx)%Cx;nxt.y = (Ay*cur.y+By)%Cy;cur = nxt;s.insert( MP(cur.x,cur.y) );n--;int64 res,Min;res = 0;Min = inf;while( n-- ){if( Min==0 ) break; nxt.x = (Ax*cur.x+Bx)%Cx;nxt.y = (Ay*cur.y+By)%Cy;cur = nxt;if( s.count(MP(cur.x,cur.y)) ) {res += 0;break;}s.insert( MP(cur.x,cur.y) );set<PII>::iterator it,tmp;it = s.lower_bound( MP(cur.x,cur.y) );for( tmp=it,tmp++;tmp!=s.end();tmp++ ){int64 tx = (*tmp).first;int64 ty = (*tmp).second;if( (tx-cur.x)*(tx-cur.x)>=Min ) break;else{int64 Dis = (tx-cur.x)*(tx-cur.x)+(ty-cur.y)*(ty-cur.y);Min = min( Min,Dis );}}for( tmp=it,tmp--;it!=s.begin();tmp-- ){int64 tx = (*tmp).first;int64 ty = (*tmp).second;if( (tx-cur.x)*(tx-cur.x)>=Min ) break;else{int64 Dis = (tx-cur.x)*(tx-cur.x)+(ty-cur.y)*(ty-cur.y);Min = min( Min,Dis );}if( tmp==s.begin() ) break;}res += Min;}//printf("%lld\n",res);printf("%I64d\n",res);}return 0;}