hdu2795Billboard(線段樹，找第一個大於w的點)

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BillboardTime Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10676    Accepted Submission(s): 4728


Problem DescriptionAt 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. 
InputThere 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. 
OutputFor 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線段樹問題，給出h*w的廣告版，每一個廣告是1*w的，給出m個廣告，要每張廣告盡量在上層盡量靠左，輸出它所在的高度，如果放不下，就輸出-1這裡的問題就是h給的很大，但是 一共只有m個廣告，所以即使是一條廣告佔一條，那麼也就只需要m的高度，多餘的高度沒有用，如果m的高度的不能放下m條廣告，那麼即使h再高，也放不下，所以n應該是m和h的小值，一段儲存下它所控制的區間的最大值，只要最大值大於廣告wi，那麼第i條廣告就可以放到這個區間內。先判斷左子樹，左子樹不合適後判斷右子樹，除非是當前的wi大於整段的最大值，否則一定可以放到廣告版內，注意，在放入一個廣告之後，修改所有的父節點的最大值。   
#include <cstdio>#include <cstring>#include <algorithm>#define INF 0x3f3f3f3fusing namespace std;#define maxn 300000#define lmin 1#define rmax n#define lson l,(l+r)/2,rt<<1#define rson (r+l)/2+1,r,rt<<1|1#define root lmin,rmax,1#define now l,r,rt#define int_now int l,int r,int rtint cl[maxn<<2] , k[maxn<<2] , top ;void push_up(int_now){    cl[rt] = max( cl[rt<<1], cl[rt<<1|1] ) ;}void creat(int w,int_now){    if( l != r )    {        creat(w,lson);        creat(w,rson);        push_up(now);    }    else    {        cl[rt] = w ;        k[rt] = ++top ;    }}int update(int w,int_now){    int ans ;    if( cl[rt] < w )        return -1 ;    if( l == r && cl[rt] >= w )    {        cl[rt] -= w ;        return k[rt] ;    }    else    {        if( cl[rt<<1] >= w )            ans = update(w,lson);        else            ans = update(w,rson);        push_up(now);        return ans;    }}int main(){    int i , n , w , m ;    while(scanf("%d %d %d", &n, &w, &m)!=EOF)    {        top = 0 ;        n = min(n,m);        creat(w,root);        while(m--)        {            scanf("%d", &w);            printf("%d\n", update(w,root));        }    }}