AtCoder Beginner Contest 259 - D - Circumferences - 题解

Posted 2022-07-13 Tisfy

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $400$ points

Problem Statement

You are given $N$ circles on the $xy$-coordinate plane. For each $i=1,2,\dots ,N$, the $i$-th circle is centered at (xi,yi)(x_i, y_i)(xi​,yi​) and has a radius of rir_iri​.

Determine whether it is possible to get from (sx,sy)(s_x, s_y)(sx​,sy​) to (tx,ty)(t_x, t_y)(tx​,ty​) by only passing through points that lie on the circumference of at least one of the $N$ circles.

Constraints

• $1\le N\le 3000$
• −109≤xi,yi≤109-10^9 \\leq x_i, y_i \\leq 10^9−109≤xi​,yi​≤109
• 1≤ri≤1091 \\leq r_i \\leq 10^91≤ri​≤109
• (sx,sy)(s_x, s_y)(sx​,sy​) lies on the circumference of at least one of the $N$ circles.
• (tx,ty)(t_x, t_y)(tx​,ty​) lies on the circumference of at least one of the $N$ circles.
• All values in input are integers.

---

Input

Input is given from Standard Input in the following format:

$N$
sxs_xsx​ sys_ysy​ txt_xtx​ tyt_yty​
x1x_1x1​ y1y_1y1​ r1r_1r1​
x2x_2x2​ y2y_2y2​ r2r_2r2​
$⋮$
xNx_NxN​ yNy_NyN​ rNr_NrN​

Output

If it is possible to get from (sx,sy)(s_x, s_y)(sx​,sy​) to (tx,ty)(t_x, t_y)(tx​,ty​), print Yes; otherwise, print No. Note that the judge is case-sensitive.

---

Sample Input 1

4
0 -2 3 3
0 0 2
2 0 2
2 3 1
-3 3 3

Sample Output 1

Yes

Here is one way to get from $(0,-2)$ to $(3,3)$.

• From $(0,-2)$, pass through the circumference of the $1$-st circle counterclockwise to reach (1,−3)(1, -\\sqrt3)(1,−3 ​).
• From (1,−3)(1, -\\sqrt3)(1,−3 ​), pass through the circumference of the $2$-nd circle clockwise to reach $(2,2)$.
• From $(2,2)$, pass through the circumference of the $3$-rd circle counterclockwise to reach $(3,3)$.

Thus, Yes should be printed.

---

Sample Input 2

3
0 1 0 3
0 0 1
0 0 2
0 0 3

Sample Output 2

No

It is impossible to get from $(0,1)$ to $(0,3)$ by only passing through points on the circumference of at least one of the circles, so No should be printed.

题目大意

给你一些圆⚪，以及两个点📍

问你 在只经过圆周⚪的前提下，能否由第一个点📍达到第二个点📍

数据保证两个点都在圆周⚪上

解题思路

本题圆⚪的数量级别为$3000$，$O(n^2)$的复杂度在AtCoder上2秒可以通过。

所以不难想到，我们可以把此题转换为连通图问题：

把一个圆⚪看成一个节点，相交的两个圆⚪之间存在一条路径

很容易在$O(n^2)$的时间内把图构建出来

然后，只需要看两个点📍所在的节点（如果某个点在多个圆⚪上，任取一个作为这个点📍所在的节点即可）是否在一个连通图上

定义圆⚪的数据结构

struct circle 
    ll x, y, r;
    bool used = false;  // 后面在判断连通图的时候会用到
;

判断两个圆⚪是否相交/相切

相交/相切 条件：$R-r\le 圆心距离\le R+r$

严格地说“相切”不属于“相交”。这里感谢@ZZXzzx0_0大佬的指正~

inline bool intersect(int x, int y) 
    ll r = a[x].r;
    ll R = a[y].r;
    if (r > R)
        swap(r, R);
    ll distance2 = (a[x].x - a[y].x) * (a[x].x - a[y].x) + (a[x].y - a[y].y) * (a[x].y - a[y].y);
    return (R - r) * (R - r) <= distance2 && distance2 <= (R + r) * (R + r);

判断一个点📍是否在一个圆⚪上

// (x, y) 是否在第th个圆⚪上
inline bool onCircle(ll x, ll y, int th) 
    return (x - a[th].x) * (x - a[th].x) + (y - a[th].y) * (y - a[th].y) == a[th].r * a[th].r;

是否能从节点x走到节点y

建好图后，假设两个点📍分别在节点$x$和节点$y$上，则只需要判断$x$和$y$是否在一个连通图上即可

bool ifCanGo(int x, int y) 
    // 把从x能到达的所有节点的used标记为true
    queue<int> canGo;
    canGo.push(x);
    a[x].used = true;
    while (canGo.size()) 
        int thisNode = canGo.front();
        canGo.pop();
        for (int &t : ma[thisNode]) 
            if (!a[t].used) 
                canGo.push(t);
                a[t].used = true;
            
        
    
    // 判断y是否被标记了
    return a[y].used;

建图

vector<int> ma[3010];  // 存图

for (int i = 0; i < n; i++) 
    for (int j = i + 1; j < n; j++) 
        if (intersect(i, j)) 
            ma[i].push_back(j);
            ma[j].push_back(i);
        
    

---

AC代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;

// 小心精度误差

// #define Yes puts("Yes"); return 0;
// #define No puts("No"); return 0;

#define EXIT(x) puts(#x); return 0;
#define Yes
#define No

struct circle 
    ll x, y, r;
    bool used = false;  
;

circle a[3010];

inline bool onCircle(ll x, ll y, int th) 
    return (x - a[th].x) * (x - a[th].x) + (y - a[th].y) * (y - a[th].y) == a[th].r * a[th].r;


inline bool intersect(int x, int y) 
    ll r = a[x].r;
    ll R = a[y].r;
    if (r > R)
        swap(r, R);
    ll distance2 = (a[x].x - a[y].x) * (a[x].x - a[y].x) + (a[x].y - a[y].y) * (a[x].y - a[y].y);
    return (R - r) * (R - r) <= distance2 && distance2 <= (R + r) * (R + r);


vector<int> ma[3010];  // 图

bool ifCanGo(int x, int y) 
    queue<int> canGo;
    canGo.push(x);
    a[x].used = true;
    while (canGo.size()) 
        int thisNode = canGo.front();
        canGo.pop();
        for<
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