hdu 2795 线段数

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3757    Accepted Submission(s): 1779

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

 

---

看完题目，我完全没有思路，因为这道题目是放在DIY里面的一个线段数和树状数组的栏目里面的，我尽量把它往线段数上靠拢，可是还是不知道怎么做，上网搜了一下，只要想通了就很简单了。

思路参照这个博客 http://blog.csdn.net/xjwjava/article/details/6387089；

代码：

#include <iostream>

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

using namespace std;



const int M=200005;     //由于n 只有200000,所以不必担心数值太大。



struct    Node{

    int left,right;

    int v;

}tree[M*3];

int ans;



int max(int a, int b)

{

    if(a>=b) return a;

    return b;

}



void build(int s, int t,int x, int r)

{

    tree[r].left=s;

    tree[r].right=t;

    tree[r].v=x;

    int mid=(tree[r].left+tree[r].right)/2;

    if(tree[r].left==tree[r].right) return;

    build(s,mid,x,r*2);

    build(mid+1,t,x,r*2+1);

}



int    query(int s, int t, int x, int r)

{

    if(tree[r].v<x)

        return 0;

    if(tree[r].left==tree[r].right && tree[r].v>=x)

    {

        tree[r].v-=x;

        ans=tree[r].left;

        return tree[r].v;

    }

    int mid=(tree[r].left+tree[r].right)/2;

    if(tree[r*2].v>=x)

    {

        return tree[r].v=max(query(s,mid,x,r*2),tree[r*2+1].v);   // 选左右两个子树中v较大的那个。

    }

    else

    {

        return tree[r].v=max(query(mid+1,t,x,r*2+1),tree[r*2].v);

    }

}



int main()

{

    int h,w,n;

    while(scanf("%d%d%d",&h,&w,&n)!=EOF)

    {

        if(h>n) h=n;

        build(1,h,w,1);

        while(n--)

        {

            int v;

            ans=-1;

            scanf("%d",&v);

            query(1,h,v,1);

            printf("%d\n",ans);



        }

    }

    return 0;

}