http://acm.hdu.edu.cn/showproblem.php?pid=4631
比赛的时候搜到了最接近点对的求法 Nlog(N） 又估摸着依次插入求的话会TLE 想了想觉得可以先把最近的位置求出来 然后后面的直接不用求了 依次直到减完 又觉得可能会有变态的数据每次最近的都在最后面 没敢写。。后来 发现它出现在题解的方法三中。。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 using namespace std; 7 #define N 500005 8 #define LL long long 9 #define INF 1000000000000000000 10 int mm; 11 LL d; 12 struct Point 13 { 14 LL x; 15 LL y; 16 int p; 17 }point[N],tt[N]; 18 int tmpt[N]; 19 20 bool cmpxy(const Point& a, const Point& b) 21 { 22 if(a.x != b.x) 23 return a.x < b.x; 24 return a.y < b.y; 25 } 26 bool cmpy(const int& a, const int& b) 27 { 28 return point[a].y < point[b].y; 29 } 30 31 LL min(LL a, LL b) 32 { 33 return a < b ? a : b; 34 } 35 36 LL dis(int i, int j) 37 { 38 return (point[i].x-point[j].x)*(point[i].x-point[j].x) 39 + (point[i].y-point[j].y)*(point[i].y-point[j].y); 40 } 41 void Closest_Pair(int left, int right) 42 { 43 if(left==right) 44 { 45 return ; 46 } 47 if(left + 1 == right) 48 { 49 LL xx = dis(left, right); 50 if(d>xx) 51 { 52 53 d = xx; 54 mm = max(point[left].p,point[right].p); 55 } 56 return ; 57 } 58 int mid = (left+right)>>1; 59 Closest_Pair(left,mid); 60 Closest_Pair(mid+1,right); 61 int i,j,k=0; 62 for(i = left; i <= right; i++) 63 { 64 if((point[mid].x-point[i].x)*(point[mid].x-point[i].x) <= d) 65 tmpt[k++] = i; 66 } 67 sort(tmpt,tmpt+k,cmpy); 68 for(i = 0; i < k; i++) 69 { 70 for(j = i+1; j < k && (point[tmpt[j]].y-point[tmpt[i]].y)*(point[tmpt[j]].y-point[tmpt[i]].y)<d; j++) 71 { 72 LL d3 = dis(tmpt[i],tmpt[j]); 73 if(d > d3) 74 { 75 d = d3; 76 mm = max(point[tmpt[j]].p,point[tmpt[i]].p); 77 } 78 } 79 } 80 } 81 int main() 82 { 83 int t,a[10],i; 84 cin>>t; 85 while(t--) 86 { 87 for(i = 1; i <= 7 ; i++) 88 cin>>a[i]; 89 point[i].x = 0; 90 point[i].y = 0; 91 for(i = 1; i <= a[1] ; i++) 92 { 93 point[i].x = (point[i-1].x*a[2]+a[3])%a[4]; 94 point[i].y = (point[i-1].y*a[5]+a[6])%a[7]; 95 point[i].p = i; 96 tt[i].x = point[i].x; 97 tt[i].y = point[i].y; 98 tt[i].p = i; 99 } 100 LL s=0; 101 int n = a[1]; 102 while(1) 103 { 104 sort(point+1,point+n+1,cmpxy); 105 d = INF; 106 Closest_Pair(1,n); 107 s+=(n-mm+1)*d; 108 n = mm-1; 109 for(i = 1 ; i <= n ; i++) 110 { 111 point[i].x = tt[i].x; 112 point[i].y = tt[i].y; 113 point[i].p = tt[i].p; 114 } 115 if(n<=1) 116 break; 117 } 118 cout<<s<<endl; 119 } 120 return 0; 121 }