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  • APIO 2018选圆圈

    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    #include<iostream>
    #include<queue>
    #include<cmath>
    #define sqr(x) (x) * (x)
    #define M 300005
    using namespace std;
    const double inf = 1e20, eps = 1e-3, alpha = acos(-1) / 5, cosa = cos(alpha), sina = sin(alpha);
    int n, root, ans[M], D,lc[M], rc[M], id[M];
    double maxx[M][2], minn[M][2], deed[M][2], rn[M];
    int read() {
        int nm = 0, f  =1;
        char c = getchar();
        for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
        for(; isdigit(c); c = getchar()) nm = nm * 10 + c  - '0';
        return nm * f;
    }
    
    struct C {
        double d[2], r;
        int id;
        bool operator < (C b) const {
            return d[D] < b.d[D];
        }
        double& operator [](int x) {
            return d[x];
        }
    
    } note[M];
    
    bool cmp(C a, C b) {
        return a.r == b.r ? a.id < b.id : a.r > b.r;
    }
    
    void updata(int x) {
        for(int i = 0; i <= 1; i++) {
            maxx[x][i] = minn[x][i] = deed[x][i];
            maxx[x][i] = max(maxx[x][i], max(maxx[lc[x]][i], maxx[rc[x]][i]));
            minn[x][i] = min(minn[x][i], min(minn[lc[x]][i], minn[rc[x]][i]));
        }
    }
    
    int build(int l, int r, int kx) {
        if(l > r) return 0;
        int mid = (l + r) >> 1;
        D = kx;
        nth_element(note + l, note + mid, note + r + 1);
        deed[mid][0] = note[mid][0], deed[mid][1] = note[mid][1];
        id[mid] = note[mid].id;
        rn[mid] = note[mid].r;
        lc[mid] = build(l, mid - 1, kx ^ 1), rc[mid] = build(mid + 1, r, kx ^ 1);
        updata(mid);
        return mid;
    }
    bool check(int x, C a)
    {
        double xn = deed[x][0], yn = deed[x][1], xnn = a[0], ynn = a[1], r1 = rn[x], r2 = a.r;
    
        return sqr(xn - xnn) + sqr(yn - ynn) - sqr(r1 + r2)<= eps; 
    }
    bool cut(int x, C a) {
        double xn = a[0], yn = a[1], r = a.r + a.r; // 位被选中的圆形大小一定比当前小  依据这个来剪枝
        if(xn + r + eps < minn[x][0]) return true;
        if(yn + r + eps < minn[x][1]) return true;
        if(xn - r - eps > maxx[x][0]) return true;
        if(yn - r - eps > maxx[x][1]) return true;
        return false;
    }
    void modify(int x, C a) {
        if(x == 0) return;
        if(cut(x, a)) return;
        if(!ans[id[x]] && check(x, a)) ans[id[x]] = a.id;
        modify(lc[x], a) ,modify(rc[x], a);
    }
    
    int main() {
        maxx[0][1] = maxx[0][0] = -inf;
        minn[0][1] = minn[0][0] = inf;
        n = read();
        for(int i = 1; i <= n; i++) {
            double x = read(), y = read(), z = read();
            note[i] = (C) {
                x * cosa - y * sina, x * sina + y * cosa, z, i
            };
        }
        root = build(1, n, 0);
        sort(note + 1, note + n + 1, cmp);
        for(int i = 1; i <= n; i++)
            if(!ans[note[i].id]) modify(root, note[i]);
        for(int i = 1; i <= n; i++) printf("%d ", ans[i]);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/luoyibujue/p/9248779.html
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