Gym 101480I Ice Igloos(思维乱搞)题解
Posted kirinsb
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了Gym 101480I Ice Igloos(思维乱搞)题解相关的知识,希望对你有一定的参考价值。
题意:给个最多500 * 500的平面,有半径最多不为1的n个圆,现在给你1e5条线段,问你每条线段和几个圆相交,时限10s
思路:
因为半径<1,那么我其实搜索的范围只要在线段附近就好了。x1 == x2 或者 y1 == y2这个很好理解,不解释。如果是斜率> 0的,那么对于任意的x (x1 <= x < x2),那我的范围就是floor(y)~ceil(y + 1),另一种斜率同理。然后我去数每一个格子有没有圆,能不能碰到我线段就行了。每个格子数完标记一下。可以偷个懒,标记为p,然后每次判是不是p,这样就省了每次都初始化。
题解虽然说最好规避sqrt,不过好像精度影响不大。
毒瘤的是输入,x1 > x2,y1 > y2。
代码:
#include<cmath> #include<set> #include<map> #include<queue> #include<cstdio> #include<vector> #include<cstring> #include <iostream> #include<algorithm> using namespace std; typedef long long ll; typedef unsigned long long ull; const int maxn = 500 + 10; const int M = maxn * 30; const ull seed = 131; const int INF = 0x3f3f3f3f; const int MOD = 1000000007; double r[maxn][maxn], k; int vis[maxn][maxn]; double a, b; double getY(double x){ return k * (x - a) + b; } bool check(double x, double y, double R){ double dis = fabs(k * x - k * a + b - y) / sqrt(k * k + 1); if(dis <= R) return true; return false; } int main(){ int n; int x1, y1, x2, y2, ans; memset(r, 0, sizeof(r)); memset(vis, 0, sizeof(vis)); scanf("%d", &n); for(int i = 1; i <= n; i++){ int u, v; double l; scanf("%d%d%lf", &u, &v, &l); r[u][v] = l; } scanf("%d", &n); for(int p = 1; p <= n; p++){ int up, down; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); a = x1, b = y1; ans = 0; k = ((double)(y2 - y1)) / ((double)(x2 - x1)); if(x1 == x2){ if(y1 > y2) swap(y1, y2); for(int i = y1; i <= y2; i++) if(r[x1][i] != 0) ans++; } else if(y1 == y2){ if(x1 > x2) swap(x1, x2); for(int i = x1; i <= x2; i++) if(r[i][y1] != 0) ans++; } else{ if(x1 > x2){ swap(x1, x2); swap(y1, y2); } if(k > 0){ for(int i = x1; i < x2; i++){ up = (int)ceil(getY(i + 1)); down = (int)floor(getY(i)); if(i == x1) down = y1; if(i == x2 - 1) up = y2; for(int j = down; j <= up; j++){ if(r[i][j] > 0 && vis[i][j] != p && check(i, j, r[i][j])){ vis[i][j] = p; ans++; } if(r[i + 1][j] > 0 && vis[i + 1][j] != p && check(i + 1, j, r[i + 1][j])){ vis[i + 1][j] = p; ans++; } } } } else{ for(int i = x1; i < x2; i++){ up = (int)ceil(getY(i)); down = (int)floor(getY(i + 1)); if(i == x1) up = y1; if(i == x2 - 1) down = y2; for(int j = down; j <= up; j++){ if(r[i][j] > 0 && vis[i][j] != p && check(i, j, r[i][j])){ vis[i][j] = p; ans++; } if(r[i + 1][j] > 0 && vis[i + 1][j] != p && check(i + 1, j, r[i + 1][j])){ vis[i + 1][j] = p; ans++; } } } } } printf("%d\\n", ans); } return 0; }
以上是关于Gym 101480I Ice Igloos(思维乱搞)题解的主要内容,如果未能解决你的问题,请参考以下文章
Ice cream samples Gym - 101670G 滑动扫描
The 2016 ACM-ICPC Asia China-Final Contest Gym - 101194D Ice Cream Tower