Gym 101480I Ice Igloos（思维乱搞）题解

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;
}

原文地址：https://www.cnblogs.com/KirinSB/p/10829107.html