Billboard
Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 24505 Accepted Submission(s): 10069
Problem Description
At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions, changes 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.
Input
There are multiple cases (no more than 40 cases).
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.
Output
For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h, starting with the top row. If an announcement can't be put on the billboard, output "-1" for this announcement.
Sample Input
3 5 5
2
4
3
3
3
Sample Output
1
2
1
3
-1
Author
hhanger@zju
Source
HDOJ 2009 Summer Exercise（5）
Recommend
lcy
题意：有一块h*w的墙，有N个宣传单 每个的大小为 1*wi 每次贴都从最左最上能贴的位置开始。求每一宣传单被贴在哪一行。
思路：以1～h划分区间，每一个区间中保存该区间能放下的最大长度，每次从最上面开始找，找到符合的位置 贴上，该位置减去len，更新区间的最大长度值。
代码：
#include <cstdio>
#include <algorithm>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
const int maxn = 222222;
int h , w , n;
int MAX[maxn<<2];
void PushUP(int rt) {
MAX[rt] = max(MAX[rt<<1] , MAX[rt<<1|1]);
}
void build(int l,int r,int rt) {
MAX[rt] = w;
if (l == r) return ;
int m = (l + r) >> 1;
build(lson);
build(rson);
}
int query(int x,int l,int r,int rt) {
if (l == r) {//找到叶子节点
MAX[rt] -= x;
return l;
}
int m = (l + r) >> 1;
int ret = (MAX[rt<<1] >= x) ? query(x , lson) : query(x , rson);
PushUP(rt);//回溯更新父亲节点
return ret;
}
int main() {
while (~scanf("%d%d%d",&h,&w,&n)) {
if (h > n) h = n;
build(1 , h , 1);
while (n --) {
int x;
scanf("%d",&x);
if (MAX[1] < x) puts("-1");
else printf("%d\n",query(x , 1 , h , 1));
}
}
return 0;
}
