Billboard(线段树)

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 524333

 

Sample Output

1213-1

 题意：

要在一块高度为h，宽度为w的广告牌上贴n条广告，每条广告的高度为1，宽度为Wi，

贴广告的人总是会优先选择最上面的位置来帖，而且在所有最上面的可能位置中，

他会选择最左面的位置，而且不能把已经贴好的广告盖住。

对每张广告，输出它被贴在的行编号（是1到h之间的数），顶部是第一行。

如果某广告不能被贴上，则输出-1。

分析：

此题对高度h建立线段树，但是h的范围是0---10^9，tree数组不能建立，

再看看n的范围0----2*10^5，n为要贴的广告数量，如果每行只贴1条广告

需要的最大高度也就是n，因此在建树时取n和h的最小值

代码：

#include<stdio.h>#include<string.h>#include<algorithm>#define MAX 200002using namespace std;struct node{    int val; //记录广告牌剩余的最大宽度}tree[MAX*4];int h,w,n;/*功能：以广告牌的纵坐标建树root：当前根结点[left,right]：当前根结点表示的范围*/void build(int root,int left,int right,int w){    tree[root].val=w;    if(left==right) return;    int mid=(left+right)/2;    build(root*2,left,mid,w);    build(root*2+1,mid+1,right,w);}/*功能：更新结点root：当前根结点[left,right]：当前根结点表示的范围w：广告的宽度*/void update(int root,int left,int right,int w){    if(left==right){ //到达叶结点        tree[root].val-=w;//剩余宽度减w        printf("%d\n",left);        return;    }    int mid=(left+right)/2;    if(w<=tree[root*2].val) update(root*2,left,mid,w); //优先贴左边    else update(root*2+1,mid+1,right,w);    tree[root].val=max(tree[root*2].val,tree[root*2+1].val);//取左右子树的最大值}int main(){    while(scanf("%d%d%d",&h,&w,&n)!=EOF){        memset(tree,0,sizeof(tree));        if(n<h) h=n;        build(1,1,h,w);//在tree中 1 号结点的根结点        int i;        for(i=0;i<n;i++){           int ww;           scanf("%d",&ww);           if(ww<=tree[1].val) update(1,1,h,ww); //根结点的值是剩余宽度的最大值           else printf("-1\n");        }    }return 0;}