Billboard
Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11900 Accepted Submission(s): 5240
in the dining room menu, and other important information.
On September 1, the billboard was empty. One by one, the announcements started being put on the billboard.
Each announcement is a stripe of paper of unit height. More specifically, the i-th announcement is a rectangle of size 1 * wi.
When someone puts a new announcement on the billboard, she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.
If there is no valid location for a new announcement, it is not put on the billboard (that's why some programming contests have no participants from this university).
Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed.
The first line of the input file contains three integer numbers, h, w, and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements.
Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement.
be put on the billboard, output "-1" for this announcement.
3 5 5 2 4 3 3 3
1 2 1 3 -1
线段树,维护一个每一行剩余长度的最大值,按高度从高到低,从左往右建树,查询时尽量往左查询
#include <map> #include <set> #include <list> #include <queue> #include <stack> #include <vector> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int N = 200010; int h, w, cnt, res; struct node { int l, r; int wide;//维护每一行剩余的最大值 int ret; }tree[N << 2]; void build(int p, int l, int r) { tree[p].l = l; tree[p].r = r; if (l == r) { tree[p].wide = w; tree[p].ret = cnt++; return; } int mid = (l + r) >> 1; build(p << 1, l, mid); build(p << 1 | 1, mid + 1, r); tree[p].wide = max(tree[p << 1].wide ,tree[p <<1 | 1].wide); } void query(int p, int ans) { if (tree[p].l == tree[p].r) { tree[p].wide -= ans; res = tree[p].ret; return; } if (tree[p << 1].wide >= ans) { query(p << 1, ans); } else { query(p << 1 | 1, ans); } tree[p].wide = max(tree[p << 1].wide, tree[p << 1 | 1].wide); } int main() { int n, ww; while (~scanf("%d%d%d", &h, &w, &n)) { cnt = 1; build(1, 1, min(h, n)); for (int i = 0; i < n; ++i) { scanf("%d", &ww); int d = tree[1].wide; if (d < ww) { printf("-1\n"); } else { query(1, ww); printf("%d\n", res); } } } return 0; }