Billboard

Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12812 Accepted Submission(s): 5578


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 | We have carefully selected several similar problems for you: 1698 1542 1828 1540 2871


題意： 有一個h*w的木板，要在上面貼廣告，有一個要求就是盡可能的貼在上面，如果不能就盡可能的貼在左面，（這樣可以使得貼的廣告數目最多吧），每個廣告都是1*wi，這就說明每一個廣告只能占據一行，可以使用線段樹進行維護，每次都進行比較，如果最上面剩下的寬度大於或等於廣告的寬就減去廣告在木板該行的寬度。如果能想起來這種方法就很簡單，如果想不出來都不知道怎麼做...

#include
#include
#include
#include
#include
#include
#include

using namespace std;

const int maxn = 200001;

struct node
{
    int l;
    int r;
    int cnt;
}q[maxn<<4];
int h,w,n;

void build(int l,int r,int rt)
{
    q[rt].l = l;
    q[rt].r = r;
    q[rt].cnt = 0;
    if(q[rt].l == q[rt].r)
    {
        q[rt].cnt = w;
        return ;
    }
    int mid = (l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    q[rt].cnt = max(q[rt<<1].cnt,q[rt<<1|1].cnt);
}

int qurry(int k,int l,int r,int rt)
{
    if(q[rt].l == q[rt].r)
    {
        q[rt].cnt -= k;
        return q[rt].l;
    }
    int mid = (l+r)>>1;
    int ans = 0;
    if(q[rt<<1].cnt>=k)
    {
        ans = qurry(k,l,mid,rt<<1);
    }
    else
    {
        ans = qurry(k,mid+1,r,rt<<1|1);
    }
    q[rt].cnt = max(q[rt<<1].cnt,q[rt<<1|1].cnt);
    return ans;
}

int main()
{
    while(scanf("%d%d%d",&h,&w,&n)!=EOF)
    {
        if(h>n)
        {
            h = n;
        }
        build(1,h,1);
        while(n--)
        {
            int m;
            scanf("%d",&m);
            if(m>q[1].cnt)
            {
                printf("-1\n");
            }
            else
            {
                int ans = qurry(m,1,h,1);
                printf("%d\n",ans);
            }

        }
    }
    return 0;
}