1) add: Mark an element in the matrix. It is guaranteed that the element has not been marked before.
2) remove: Delete an element’s mark. It is guaranteed that the element has been marked before.
3) find: For a given element’s row and column, return a marked element’s row and column, where the marked element’s row and column are larger than the given element’s row and column respectively. If there are multiple solutions, return the element whose row is the smallest; and if there are still multiple solutions, return the element whose column is the smallest. If there is no solution, return -1.
LMY lets YY develop a program to solve the problem. Could you also develop a program to solve the problem?
End of input is indicated by a line containing a zero.
5 add 48 1 add 25 69 add 88 52 remove 25 69 add 23 89 10 add 47 23 find 66 83 find 27 73 add 84 97 find 10 58 remove 47 23 add 41 89 remove 41 89 find 65 68 add 25 41 0
Case 1: Case 2: -1 -1 84 97 84 97
//
#include <cstdio>
#include <cstring>
#include <set>
#include <algorithm>
using namespace std;
#define MAXN 200000
int sx[MAXN], tot;
set<int> sy[MAXN];
struct Node {
int _a, _b;
Node *_left, *_right;
int _max_y;
} nodes[MAXN * 2], *ptr;
void build_tree(int a, int b) {
Node* root = ++ptr;
root->_a = a; root->_b = b; root->_max_y = -1;
if (root->_a == root->_b) {
root->_left = root->_right = 0;
return;
}
root->_left = ptr + 1;
build_tree(a, (a + b) / 2);
root->_right = ptr + 1;
build_tree((a + b) / 2 + 1, b);
}
int lx, ly;
char lo;
void modify(Node* root) {
if (root->_a == root->_b) {
if (lo == 'a') {
sy[root->_a].insert(ly);
root->_max_y = *sy[root->_a].rbegin();
} else {
sy[root->_a].erase(ly);
if (sy[root->_a].size() == 0)
root->_max_y = -1;
else
root->_max_y = *sy[root->_a].rbegin();
}
return;
}
int mid = (root->_a + root->_b) / 2;
if (lx <= sx[mid])
modify(root->_left);
else
modify(root->_right);
root->_max_y = max(root->_left->_max_y, root->_right->_max_y);
}
int la, lb;
int query(Node* root) {
if (la <= root->_a && root->_b <= lb)
return root->_max_y;
int mid = (root->_a + root->_b) / 2, res = -1;
if (la <= mid)
res = query(root->_left);
if (mid < lb)
res = max(res, query(root->_right));
return res;
}
int get_x_index(int x, int y) {
int pos = upper_bound(sx, sx + tot, x) - sx, r = tot - 1, mid, l = pos - 1;
if (pos == tot)
return -1;
la = pos, lb = tot - 1;
if (query(nodes + 1) <= y)
return -1;
while (l + 1 != r) {
mid = (l + r) / 2;
la = pos; lb = mid;
if (query(nodes + 1) > y)
r = mid;
else
l = mid;
}
return r;
}
struct QNode {
int _x, _y;
char _opt[10];
} q[MAXN];
int main() {
int n, cas = 0;
while (scanf("%d", &n) != EOF) {
if (n == 0) break;
tot = 0;
for (int i = 0; i < n; ++i) {
scanf("%s%d%d", q[i]._opt, &q[i]._x, &q[i]._y);
if (q[i]._opt[0] == 'a')
sx[tot++] = q[i]._x;
}
sort(sx, sx + tot);
tot = unique(sx, sx + tot) - sx;
ptr = nodes;
build_tree(0, tot - 1);
if (cas)
putchar('\n');
printf("Case %d:\n", ++cas);
int idx;
for (int i = 0; i < n; ++i)
switch (q[i]._opt[0]) {
case 'a': case 'r':
lx = q[i]._x;
ly = q[i]._y;
lo = q[i]._opt[0];
modify(nodes + 1);
break;
case 'f':
idx = get_x_index(q[i]._x, q[i]._y);
if (idx == -1)
puts("-1");
else
printf("%d %d\n", sx[idx], *sy[idx].upper_bound(q[i]._y));
}
for (int i = 0; i < tot; ++i)
sy[i].clear();
}
return 0;
}
这篇博客介绍了一个在二维平面上进行点添加和删除的问题,利用离散化和线段树数据结构,高效地找到横纵坐标都大于已添加点的最小坐标。首先按横坐标选择最小的点,其次按纵坐标选择最小的点。博客中给出了详细的算法实现,包括build_tree、modify和query等函数。
1126

被折叠的 条评论
为什么被折叠?



