HDU 2795-Billboard

(题目链接)[http://acm.hdu.edu.cn/showproblem.php?pid=2795]

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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

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题目翻译：有一个h*w长度的木板，现在要向上贴n张广告，广告之间不能覆盖，且每一张广告的高度都为1，第i张广告的宽度为w[i]。并且贴广告的规则是
每次都铁道最左上方能够贴得下的位置。并输出该行行号。
解题方法：
用高度来构建线段树，记录left行到right行中的最大留白空间，然后每贴一个广告，就在线段树中插，如果左子树的留白长度大于当前要贴的广告长度，就先
查找左子树，因为题目要求贴最左上方，当找到贴的位置，则该行的新的剩余长度就在原值上减去当前长度。并更新整科线段树。

代码

#include <iostream>#include <bits/stdc++.h>#define lchild left, mid, root<<1#define rchild mid+1, right, root<<1|1using namespace std;int height, width, n;int Max[100000];void push1(int root){    Max[root] = max(Max[root<<1],Max[root<<1|1]);}//建立线段树 void buildtree(int left, int right, int root){    Max[root] = width;    if (left == right)        return;    int mid = (left+right)>>1;    buildtree(lchild);    buildtree(rchild);}int inquiry (int left, int right, int root, int w){    if (left == right)    {        Max[root] -= w;        return left;    }    int mid = (left+right)>>1;    int ans = 0;    if (w <= Max[root<<1])        ans = inquiry(lchild, w);    else        ans = inquiry(rchild, w);    push1(root);    return ans;}int main(){    while (cin>>height>>width>>n)    {        if (n<height)            height = n;        buildtree(1,height,1);        while (n--)        {            int w;  //小广告的宽度            cin>>w;            if (Max[1] < w){                cout<<"-1"<<endl;            }            else            {                int ans = inquiry(1,height,1,w);                cout<<ans<<endl;            }        }    }    return 0;}