Billboard

Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 13 Accepted Submission(s): 11

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

Author

hhanger@zju

Source

HDOJ 2009 Summer Exercise（5）

题意：

一个h*w的牌子，向上贴广告，先从上边贴，先从左边贴，给出n个广告的长度（宽都是1），问他贴在第几行。

代码：

//线段树求最值变形，初始化叶子节点存储该行剩余容量宽度，每输入一个广告，从线段树中找，
//如果左子树中有位置就找向左子树否则找向右子树，找到之后减去广告宽度，期间更新最值。
#include<iostream>
#include<cstdio>
using namespace std;
const int maxn=200005;
int Max[maxn*4];
void pushup(int rt){
    Max[rt]=max(Max[rt<<1],Max[rt<<1|1]);
}
void build(int c,int l,int r,int rt){
    Max[rt]=c;
    if(l==r) return;
    int m=(l+r)>>1;
    build(c,l,m,rt<<1);
    build(c,m+1,r,rt<<1|1);
}
int update(int c,int l,int r,int rt){
    if(l==r){
        Max[rt]-=c;
        return l;
    }
    int m=(l+r)>>1,res;
    if(Max[rt<<1]>=c) res=update(c,l,m,rt<<1);
    else res=update(c,m+1,r,rt<<1|1);
    pushup(rt);
    return res;
}
int main()
{
    int h,w,n,x;
    while(scanf("%d%d%d",&h,&w,&n)==3){
        if(h>n) h=n;//最多用到n行，h不用1e9
        build(w,1,h,1);
        for(int i=0;i<n;i++){
            scanf("%d",&x);
            if(Max[1]<x) printf("-1\n");
            else printf("%d\n",update(x,1,h,1));
        }
    }
    return 0;
}

来源：https://www.cnblogs.com/--ZHIYUAN/p/6402139.html

标签

线段树

HDU2795 线段树

Billboard