Billboard

題目描述：

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.

輸入：

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.

輸出：

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.

樣例輸入：

3 5 5
2
4
3
3
3      

樣例輸出：

1
2
1
3
-1      

大體題意是有一個廣告牌，高為h，寬為w，現在有很多廣告要往這塊廣告牌上弄，每塊廣告高都為1，寬為wi。往這塊廣告牌上弄的原則就是盡量往上面的位置放，同一行盡量往左放。把[1，h]當做一個區間，利用線段樹求解。每當某一行滿足條件，就減去它相應的寬度。

AC代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=200010;
int MAX[maxn<<2];
int h,w,n;
void pushup(int rt)
{
    MAX[rt]=max(MAX[rt<<1],MAX[rt<<1|1]);
}
void build(int l,int r,int rt)
{
    if(l==r)//初始化每一行的寬度都為w
    {
        MAX[rt]=w;
        return;
    }
    int m=(l+r)>>1;
    build(l,m,rt<<1);
    build(m+1,r,rt<<1|1);
    pushup(rt);
}
int  Query(int x,int l,int r,int rt)
{
    if(l==r)
    {
        MAX[rt]-=x;
        return l;//注意一定要傳回左值
    }
    int m=(l+r)>>1;
    int ans=MAX[rt<<1];//左子區間
    if(ans>=x)//如果左子區間能放下這塊廣告，那就更新左子區間，否則更新右子區間
        ans=Query(x,l,m,rt<<1);
    else
        ans=Query(x,m+1,r,rt<<1|1);
    pushup(rt);
    return ans;
}
int main()
{
    int x;
    while(scanf("%d%d%d",&h,&w,&n)!=EOF)
    {
        if(h>=n)//因為n快廣告最多就占n行，如果h大于n，那超出n的多餘的h就沒有用，這一步如果沒
            h=n;//有會runtime error
        build(1,h,1);
        while(n--)
        {
            scanf("%d",&x);
            if(MAX[1]<x)//如果總區間（最大的那個區間）都不能放開這個x寬度的廣告牌，那就說明
                printf("-1\n");//哪裡也放不開它了，就輸出-1
            else
                printf("%d\n",Query(x,1,h,1));
        }
    }
            return 0;
}