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裸的二维线段树。。。求矩形最大值和矩形覆盖
板子敲起来比较麻烦QAQ
1 /************************************************************** 2 Problem: 1513 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:24884 ms 7 Memory:127268 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 15 inline int read(); 16 17 int n, m; 18 19 struct segx { 20 segx *ls, *rs; 21 int mx, tag; 22 23 segx() { 24 ls = rs = NULL, mx = tag = 0; 25 } 26 27 #define Len (1 << 16) 28 inline void* operator new(size_t) { 29 static segx *mempool, *c; 30 if (c == mempool) 31 mempool = (c = new segx[Len]) + Len; 32 *c = segx(); 33 return c++; 34 } 35 #undef Len 36 37 #define mid (l + r >> 1) 38 void build(int l, int r) { 39 if (l == r) return; 40 ls = new()segx, rs = new()segx; 41 ls -> build(l, mid), rs -> build(mid + 1, r); 42 } 43 44 void modify(int l, int r, int L, int R, int v) { 45 mx = max(mx, v); 46 if (L <= l && r <= R) { 47 tag = max(tag, v); 48 return; 49 } 50 if (L <= mid) ls -> modify(l, mid, L, R, v); 51 if (mid < R) rs -> modify(mid + 1, r, L, R, v); 52 } 53 54 int query(int l, int r, int L, int R) { 55 if (L <= l && r <= R) return mx; 56 int res = tag; 57 if (L <= mid) res = max(res, ls -> query(l, mid, L, R)); 58 if (mid < R) res = max(res, rs -> query(mid + 1, r, L, R)); 59 return res; 60 } 61 #undef mid 62 }; 63 64 struct segy { 65 segy *ls, *rs; 66 segx *mx, *tag; 67 68 segy() { 69 ls = rs = NULL, mx = tag = NULL; 70 } 71 72 #define Len (1 << 16) 73 inline void* operator new(size_t) { 74 static segy *mempool, *c; 75 if (c == mempool) 76 mempool = (c = new segy[Len]) + Len; 77 *c = segy(); 78 return c++; 79 } 80 #undef Len 81 82 #define mid (l + r >> 1) 83 void build(int l, int r) { 84 mx = new()segx, tag = new()segx; 85 mx -> build(1, m), tag -> build(1, m); 86 if (l == r) return; 87 ls = new()segy, rs = new()segy; 88 ls -> build(l, mid), rs -> build(mid + 1, r); 89 } 90 91 void modify(int l, int r, int L, int R, int x, int y, int v) { 92 mx -> modify(1, m, x, y, v); 93 if (L <= l && r <= R) { 94 tag -> modify(1, m, x, y, v); 95 return; 96 } 97 if (L <= mid) ls -> modify(l, mid, L, R, x, y, v); 98 if (mid < R) rs -> modify(mid + 1, r, L, R, x, y, v); 99 } 100 101 int query(int l, int r, int L, int R, int x, int y) { 102 if (L <= l && r <= R) return mx -> query(1, m, x, y); 103 int res = tag -> query(1, m, x, y); 104 if (L <= mid) res = max(res, ls -> query(l, mid, L, R, x, y)); 105 if (mid < R) res = max(res, rs -> query(mid + 1, r, L, R, x, y)); 106 return res; 107 } 108 #undef mid 109 } *T; 110 111 int main() { 112 int x, y, dx, dy, h, Tot; 113 n = read(), m = read(), Tot = read(); 114 T = new()segy; 115 T -> build(1, n); 116 while (Tot--) { 117 dx = read(), dy = read(), h = read(), x = read(), y = read(); 118 T -> modify(1, n, x + 1, x + dx, y + 1, y + dy, T -> query(1, n, x + 1, x + dx, y + 1, y + dy) + h); 119 } 120 printf("%d\n", T -> query(1, n, 1, n, 1, m)); 121 return 0; 122 } 123 124 inline int read() { 125 static int x; 126 static char ch; 127 x = 0, ch = getchar(); 128 while (ch < ‘0‘ || ‘9‘ < ch) 129 ch = getchar(); 130 while (‘0‘ <= ch && ch <= ‘9‘) { 131 x = x * 10 + ch - ‘0‘; 132 ch = getchar(); 133 } 134 return x; 135 }
BZOJ1513 [POI2006]Tet-Tetris 3D
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原文地址:http://www.cnblogs.com/rausen/p/4474817.html