HDU 2795 Billboard（线段树）

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

Billboard

Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20210 Accepted Submission(s): 8403

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

【中文题意】给你一个h*w的木板（长为h，宽为w），然后让你往上面贴1 *wi的海报，贴的规则是从最左而且最上开始贴，如果不能贴下的话，输出-1，如果能贴下，输出开始贴的位置。
【思路分析】用线段树来维护海报板，然后进行单点更新。
【AC代码】

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int h,w,n;struct Seg{    int left,right,value;}c[600005];void build_tree(int l,int r,int root){    c[root].left=l;    c[root].right=r;    if(l==r)    {        c[root].value=w;        return ;    }    int mid=(c[root].left+c[root].right)/2;    build_tree(l,mid,root*2);    build_tree(mid+1,r,root*2+1);    c[root].value=max(c[root*2].value,c[root*2+1].value);}int query(int l,int r,int root,int x){    if(l==r)    {        c[root].value-=x;        return l;    }    int mid=(c[root].left+c[root].right)/2;    int re;    if(c[root*2].value>=x)//如果左边能贴上去，贴在左边，否则贴在右边    {        re=query(l,mid,root*2,x);    }    else    {        re=query(mid+1,r,root*2+1,x);    }    c[root].value=max(c[root*2].value,c[root*2+1].value);//更新最大值    return re;}int main(){    while(~scanf("%d%d%d",&h,&w,&n))    {        if(h>n)//没有必要开那么大的线段树，因为每一份海报的长度都为1            h=n;        memset(c,0,sizeof(c));        build_tree(1,h,1);        int num;        for(int i=0;i<n;i++)        {             scanf("%d",&num);             if(c[1].value<num)//判断一下还能不能贴上去             printf("-1\n");             else                printf("%d\n",query(1,h,1,num));        }    }    return 0;}