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 }
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posted on 2015-05-03 21:59  Xs酱~  阅读(279)  评论(0编辑  收藏  举报