题解

二分一个横坐标，过这个横坐标做一条和y轴平行的直线，相当于在这条直线上做区间覆盖，如果区间有交的话，那么答案是True
否则的话取两个不相交的区间，如果这两个圆相离或相切则不合法

否则看看相交的部分在二分的横坐标的左边还是右边，进行更新

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {res = 0;char c = getchar();T f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 + c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10);
}int N,tot;
bool dcmp(db a,db b) {return fabs(a - b) < eps;
}
struct Point {db x,y;Point(db _x = 0.0,db _y = 0.0) {x = _x;y = _y;}friend Point operator + (const Point &a,const Point &b) {return Point(a.x + b.x,a.y + b.y);}friend Point operator - (const Point &a,const Point &b) {return Point(a.x - b.x,a.y - b.y);}friend db operator * (const Point &a,const Point &b) {return a.x * b.y - a.y * b.x;}friend Point operator * (const Point &a,const db &d) {return Point(a.x * d,a.y * d);}friend Point operator / (const Point &a,const db &d) {return Point(a.x / d,a.y / d);}friend db dot(const Point &a,const Point &b) {return a.x * b.x + a.y * b.y;}db norm() {return x * x + y * y;}};
struct Circle {Point O;db R;
}C[MAXN];
struct line {db s,t;int id;friend bool operator < (const line &a,const line &b) {return a.t < b.t || (a.t == b.t && a.s < b.s);}
}L[MAXN];
db dis(Point a,Point b) {return sqrt((b - a).norm());
}
int check(db mid) {tot = 0;for(int i = 1 ; i <= N ; ++i) {if(fabs(mid - C[i].O.x) >= C[i].R) {if(mid > C[i].O.x) return -1;else return 1;}db t = sqrt(C[i].R * C[i].R - (C[i].O.x - mid) * (C[i].O.x - mid));L[++tot] = (line){C[i].O.y - t,C[i].O.y + t,i};}sort(L + 1,L + tot + 1);db a = L[1].s,b = L[1].t;for(int i = 2 ; i <= N ; ++i) {a = max(a,L[i].s);b = min(b,L[i].t);}if(a + eps < b) return 0;for(int i = 2 ; i <= N ; ++i) {if(L[i].s >= L[1].t) {int u = L[1].id,v = L[i].id;if(dis(C[u].O,C[v].O) >= C[u].R + C[v].R) return -2;else {Point p = C[u].O + (C[v].O - C[u].O) * (C[u].R / dis(C[v].O,C[u].O));if(p.x < mid) return -1;else return 1;}} }
}
void Solve() {read(N);for(int i = 1 ; i <= N ; ++i) {scanf("%lf%lf%lf",&C[i].O.x,&C[i].O.y,&C[i].R);}db L = C[1].O.x - C[1].R,R = C[1].O.x + C[1].R;for(int i = 2 ; i <= N ; ++i) {L = min(L,C[i].O.x - C[i].R);R = max(R,C[i].O.x + C[i].R);}int cnt = 50;while(cnt--) {db mid = (L + R) / 2;int x = check(mid);if(x == -2) {puts("NO");return;}if(x == 0) {puts("YES");return;}if(x == -1) {R = mid;}else {L = mid;}}if(check(L) != 0) puts("NO");else puts("YES");
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifSolve();return 0;
}