标签:acm 线段树
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Billboard
Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11433 Accepted Submission(s): 5036
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
Sample Output
题意:
给出一个H*W的宣传板 贴高为1 长为w[i]的宣传单 尽量往上贴 问这张宣传单最高能贴在哪里 如果不能 输出-1
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=222222;
int Max[MAXN<<2];
void pushup(int rt)
{
Max[rt]=max(Max[rt<<1],Max[rt<<1|1]);
}
int query(int a,int l,int r,int rt)
{
if(l==r)
{
Max[rt]-=a;
return l;
}
int mid=(l+r)>>1;
int cnt;
if(Max[rt<<1]>=a)
cnt=query(a,l,mid,rt<<1);
else cnt=query(a,mid+1,r,rt<<1|1);
pushup(rt);
return cnt;
}
int main()
{
int w,h,n;
while(scanf("%d %d %d",&h,&w,&n)!=EOF)
{
if(n<=h)
h=n;
for(int i=0;i<=h<<2;i++)
Max[i]=w;
int a;
for(int i=0;i<n;i++)
{
scanf("%d",&a);
if(a>Max[1])
printf("-1\n");
else
{
printf("%d\n",query(a,1,h,1));
}
}
}
return 0;
}
HDU 2795 Billboard 线段树
标签:acm 线段树
原文地址:http://blog.csdn.net/qq_16843991/article/details/39522881