hdu 2795 Billboard（线段树之 单点更新+区间最值）

Billboard

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

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
   
   
   
   

 

题意：给出一个h*w的广告牌，行号从上到下依次为1,2，……，h，每次往里面贴一张1*Wi的小广告，小广告尽量贴在上面，在上面的尽量贴在左边。求每张广告应贴在哪一行，输出行号；如果没有地方贴，输出"-1"。

分析：刚拿到这个题时，没有想到用线段树来做。后来看了一下大牛的博客，才恍然大悟。因为每张广告高度为1，所以我们可以用广告牌的高度来建立一棵线段树。但是h<=10^9，叶子节点太多；而n<=200000，所以我们可以用min(h, w)来作为叶子节点的个数进行建树出。记录每个区间的最大值，查找时只需找满足条件的最小行号。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define lson l, mid, root<<1
#define rson mid+1, r, root<<1|1
const int N = 200005;
int MAX[N<<2];
int n, h, w;

void Push_Up(int root)
{
    MAX[root] = max(MAX[root<<1], MAX[root<<1|1]);
}
void Build_Tree(int l, int r, int root)
{
    MAX[root] = w;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    Build_Tree(lson);
    Build_Tree(rson);
}
int Query(int x, int l, int r, int root)
{
    if(l == r)
    {
        MAX[root] -= x;
        return l;
    }
    int mid = (l + r) >> 1;
    int ret;
    if(MAX[root<<1] >= x)
        ret = Query(x, lson);
    else
        ret = Query(x, rson);
    Push_Up(root);
    return ret;
}

int main()
{
    int x;
    while(~scanf("%d%d%d",&h,&w, &n))
    {
        if(h > n) h = n;
        Build_Tree(1, h, 1);
        while(n--)
        {
            scanf("%d",&x);
            if(MAX[1] < x) printf("-1\n");
            else printf("%d\n", Query(x, 1, h, 1));
        }
    }
    return 0;
}