題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=2795
Billboard
Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16673 Accepted Submission(s): 7056
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)
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題目大意:一個h*w的公告牌,要在其上貼公告。
輸入的是1*wi的w值,這些是公告的尺寸接下來要滿足的條件有:1、盡量往上,同一高度盡量靠左。2、求第n個廣告所在的行數。3、沒有合適的位置貼了則輸出-1。
解題思路:利用線段樹可以得出區間的最大值,在結構體中定義一個Max變量,用來表示這段區間內有最多空位的那一行的空位長度。與輸入進來的長度進行比較,先左邊比較,再右邊。(也就是左子樹的最大值大於他,就查詢左子樹,否則查詢右子樹)。
詳見代碼。
#include
#include
#include
using namespace std;
struct node
{
int l,r;
int Max;//這段區間內有最多空位的那一行的空位長度
} s[200000*4];
void InitTree(int l,int r,int k,int w)
{
s[k].l=l;
s[k].r=r;
s[k].Max=w;
if (l==r)
return ;
int mid=(l+r)/2;
InitTree(l,mid,2*k,w);
InitTree(mid+1,r,2*k+1,w);
}
void UpdataTree(int ww,int k)
{
if (s[k].l==s[k].r)
{
printf ("%d\n",s[k].l);
s[k].Max-=ww;
return ;
}
if (ww<=s[k*2].Max)
UpdataTree(ww,k*2);
else
UpdataTree(ww,k*2+1);
s[k].Max=s[k*2].Max>s[k*2+1].Max?s[k*2].Max:s[k*2+1].Max;
}
int main()
{
int h,w,n;
int ww;
while (~scanf("%d%d%d",&h,&w,&n))
{
if (h>n)//h<=n,大了就沒有意義了
h=n;
InitTree(1,h,1,w);
for (int i=1; i<=n; i++)
{
scanf("%d",&ww);
if (s[1].Max>=ww)
UpdataTree(ww,1);
else
printf ("-1\n");
}
}
return 0;
}