米啥好讲滴。。就是StL <set>的用法。。。。
Sad Love Story
Time Limit: 40000/20000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1489 Accepted Submission(s): 466
Problem Description
There's a really sad story.It could be about love or about money.But love will vanish and money will be corroded.These points will last forever.So this time it is about points on a plane.
We have a plane that has no points at the start.
And at the time i,we add point pi(xi, yi).There is n points in total.
Every time after we add a point,we should output the square of the distance between the closest pair on the plane if there's more than one point on the plane.
As there is still some love in the problem setter's heart.The data of this problem is randomly generated.
To generate a sequence x1, x2, ..., xn,we let x0 = 0,and give you
3 parameters:A,B,C. Then xi = (xi-1 * A + B) mod C.
The parameters are chosen randomly.
To avoid large output,you simply need output the sum of all answer in one line.
We have a plane that has no points at the start.
And at the time i,we add point pi(xi, yi).There is n points in total.
Every time after we add a point,we should output the square of the distance between the closest pair on the plane if there's more than one point on the plane.
As there is still some love in the problem setter's heart.The data of this problem is randomly generated.
To generate a sequence x1, x2, ..., xn,we let x0 = 0,and give you
3 parameters:A,B,C. Then xi = (xi-1 * A + B) mod C.
The parameters are chosen randomly.
To avoid large output,you simply need output the sum of all answer in one line.
Input
The first line contains integer T.denoting the number of the test cases.
Then each T line contains 7 integers:n Ax Bx Cx Ay By
Cy.
Ax,Bx,Cx is the given parameters for x1, ..., xn.
Ay,By,Cy is the given parameters for y1, ..., yn.
T <= 10.
n <= 5 * 105.
104 <= A,B,C <= 106.
Then each T line contains 7 integers:n Ax Bx Cx Ay By
Cy.
Ax,Bx,Cx is the given parameters for x1, ..., xn.
Ay,By,Cy is the given parameters for y1, ..., yn.
T <= 10.
n <= 5 * 105.
104 <= A,B,C <= 106.
Output
For each test cases,print the answer in a line.
Sample Input
2 5 765934 377744 216263 391530 669701 475509 5 349753 887257 417257 158120 699712 268352
Sample Output
8237503125 49959926940HintIf there are two points coincide,then the distance between the closest pair is simply 0.
#include <cstdio>
#include <cstdlib>
#include <set>
#include <iostream>
using namespace std;
#define LL long long
const int MAXN=555555;
LL x[MAXN],y[MAXN];
struct node{
LL x,y;
bool const operator < (const node& tmp) const{
return x<tmp.x||(x==tmp.x&&y<tmp.y);
}
};
void calcin(int n,LL s[]){
LL a,b,c;
cin>>a>>b>>c;
s[0]=0;
for(int i=1;i<=n;i++){
s[i]=(s[i-1]*a+b)%c;
// printf("s[%d]=%lld\n",i,s[i]);
}
}
void solve(int n){
calcin(n,x);
calcin(n,y);
set<node> s;
set<node>::iterator it,t1,t2;
node tmp;
LL ans=1LL<<60,k,ret=0;
for(int i=1;i<=n;i++){
tmp.x=x[i],tmp.y=y[i];
if(i>1){
it=s.lower_bound(tmp);
for(t1=it;t1!=s.end();t1++){
k=(t1->x-tmp.x)*(t1->x-tmp.x);
if(k>ans)
break;
k+=(t1->y-tmp.y)*(t1->y-tmp.y);
ans=min(k,ans);
}
for(t2=it;t2!=s.begin();){
--t2;
k=(t2->x-tmp.x)*(t2->x-tmp.x);
if(k>=ans)
break;
k+=(t2->y-tmp.y)*(t2->y-tmp.y);
ans=min(k,ans);
}
ret+=ans;
}
s.insert(tmp);
}
cout<<ret<<endl;
}
int main(){
int t,n;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
solve(n);
}
}