先按照到站的点的距离排个序，然后贪心的每次拿当前范围最大的来找
然后集齐了一组范围递增，可控重量递减的磁铁以后，就可以直接一个个看能不能拿了
但是会TLE，用线段树维护即可
1 /************************************************************** 2 Problem: 3276 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:4884 ms 7 Memory:19364 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 typedef long long ll; 15 const int N = 250005; 16 17 struct data { 18 int w, p; 19 ll d2, r2; 20 21 inline bool operator < (const data &a) const { 22 return d2 < a.d2; 23 } 24 } a[N]; 25 26 struct seg_node { 27 int v; 28 seg_node *son[2]; 29 } *seg_root, mempool[N << 2], *cnt_seg = mempool, *null; 30 31 int n; 32 int q[N], H, T; 33 ll R2; 34 35 inline int lower() { 36 int l = 1, r = n + 1, mid; 37 while (l + 1 < r) { 38 mid = l + r >> 1; 39 if (a[mid].d2 <= R2) l = mid; 40 else r = mid; 41 } 42 return l; 43 } 44 45 inline int read() { 46 int x = 0, sgn = 1; 47 char ch = getchar(); 48 while (ch < '0' || '9' < ch) { 49 if (ch == '-') sgn = -1; 50 ch = getchar(); 51 } 52 while ('0' <= ch && ch <= '9') { 53 x = x * 10 + ch - '0'; 54 ch = getchar(); 55 } 56 return sgn * x; 57 } 58 59 #define mid (l + r >> 1) 60 #define V p -> v 61 #define Ls p -> son[0] 62 #define Rs p -> son[1] 63 inline void new_node(seg_node *&p) { 64 p = ++cnt_seg; 65 V = 0, Ls = Rs = null; 66 } 67 68 inline int get_min(int x, int y) { 69 if (!x) return y; 70 if (!y) return x; 71 return a[x].w < a[y].w ? x : y; 72 } 73 74 inline void update(seg_node *p) { 75 V = get_min(Ls -> v, Rs -> v); 76 } 77 78 void seg_build(seg_node *&p, int l, int r) { 79 new_node(p); 80 if (l == r) { 81 V = l; 82 return; 83 } 84 seg_build(Ls, l, mid), seg_build(Rs, mid + 1, r); 85 update(p); 86 } 87 88 void seg_modify(seg_node *p, int l, int r, int pos) { 89 if (l == r) { 90 V = 0; 91 return; 92 } 93 if (pos <= mid) seg_modify(Ls, l, mid, pos); 94 else seg_modify(Rs, mid + 1, r, pos); 95 update(p); 96 } 97 98 int seg_query(seg_node *p, int l, int r, int pos) { 99 if (r <= pos) return V; 100 int res = seg_query(Ls, l, mid, pos); 101 if (pos > mid) res = get_min(res, seg_query(Rs, mid + 1, r, pos)); 102 return res; 103 } 104 #undef mid 105 #undef V 106 #undef Ls 107 #undef Rs 108 109 inline ll sqr(ll x) { 110 return x * x; 111 } 112 113 int main() { 114 int X, Y, P, x, y, r, i, tmp, pos; 115 null = cnt_seg, null -> son[0] = null -> son[1] = null, null -> v = 0; 116 X = read(), Y = read(), P = read(), R2 = sqr(read()), n = read(); 117 for (i = 1; i <= n; ++i) { 118 x = read(), y = read(), a[i].w = read(), a[i].p = read(), r = read(); 119 a[i].d2 = sqr(X - x) + sqr(Y - y), a[i].r2 = sqr(r); 120 } 121 sort(a + 1, a + n + 1); 122 seg_build(seg_root, 1, n); 123 while ((pos = lower()) != 0) { 124 tmp = seg_query(seg_root, 1, n, pos); 125 if (!tmp || a[tmp].w > P) break; 126 seg_modify(seg_root, 1, n, tmp); 127 q[++T] = tmp; 128 } 129 while (H <= T) { 130 P = a[q[H]].p, R2 = a[q[H]].r2, ++H; 131 while ((pos = lower()) != 0) { 132 tmp = seg_query(seg_root, 1, n, pos); 133 if (!tmp || a[tmp].w > P) break; 134 seg_modify(seg_root, 1, n, tmp); 135 q[++T] = tmp; 136 } 137 } 138 printf("%d ", T); 139 return 0; 140 }