标签:define sqrt return name abs == 分享 排序 flag
题目大意:给你n个哥布林的坐标,和m个圆,问有多少哥布林不在圆内?
标解就是:扫描线
我们可以将一个圆划分成 2*r + 1部分,然后我们对每一部分求出其上界和下界的左边,并分别打上 上界和下界的标记,然后我们将其加入哥布林那个集合,并按照 x从小到大,y从小到大 进行排序,这样我们就可以在O(n)内确定哪些哥布林在圆中(在上界和下界之间的哥布林就是在圆中的),哪些不在圆中。下面就是代码:
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define IN 0 5 #define OUT 2 6 #define GOBLIN 1 7 8 typedef long long LL; 9 const int INF = 0x3f3f3f3f; 10 const int maxn = 10000000 + 10; 11 const double eps = 1e-5; 12 13 struct Goblins{ 14 int x, t; 15 double y; 16 Goblins(int x=0, double y=0.0, int t=GOBLIN):x(x),y(y),t(t){} 17 bool operator < (const Goblins &tmp) const { 18 if(x != tmp.x) return x < tmp.x; 19 else if(fabs(y-tmp.y) >= eps) return y < tmp.y; 20 else return t < tmp.t; 21 } 22 }; 23 24 Goblins g[maxn]; 25 26 int main(){ 27 int n, m, all; 28 29 scanf("%d", &n); 30 for(int i = 0; i < n; ++i) 31 scanf("%d%lf", &g[i].x, &g[i].y); 32 33 scanf("%d", &m); 34 for(int i = 0; i < m; ++i) { 35 int x, r; 36 double y; 37 scanf("%d%lf%d", &x, &y, &r); 38 for(int j = x-r; j <= x+r; j++) { 39 double tmp = (double) (r*r - (x-j)*(x-j)); 40 g[n++] = Goblins(j, sqrt(tmp)+y, OUT); 41 g[n++] = Goblins(j, -sqrt(tmp)+y, IN); 42 } 43 } 44 45 int ans = 0, flag = 0; 46 sort(g, g+n); 47 for(int i = 0; i < n; ++i) { 48 if(g[i].t == IN) flag++; 49 if(g[i].t == OUT) flag--; 50 if(flag == 0 && g[i].t == GOBLIN) ans++; 51 } 52 53 printf("%d\n", ans); 54 return 0; 55 }
2015 NCPC Problem G-Goblin Garden Guards
标签:define sqrt return name abs == 分享 排序 flag
原文地址:https://www.cnblogs.com/DynastySun/p/9384718.html