Billboard HDU 2795

题目链接:点我

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    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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题意:

在一个 h * w 的黑板上贴广告,每次优先贴最上面的位置,在高度相同的情况下优先贴左边的位置,输出贴的广告所在的行数.

思路:

线段树的单点更新,将黑板看出h个点,每个点所能贴的广告长度为W,维护每个点的所能贴的最大的广告长度,主要的是h远大于你需要贴的广告数,所以当h大于n时, 我们要用n来建数.

代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;

struct ss{
    int num,w;
}a[800000];

void pushup(int cur){
    a[cur].w = max( a[cur << 1].w, a[cur << 1|1].w);//维护最大值
}

void build(int cur, int l, int r, int w){
    a[cur].w = w;
    if(l==r){
        a[cur].num = l;//当前的点的行数,即答案,
        return ;
    }
    int mid = (l + r) >> 1;
    build( cur << 1, l, mid, w);
    build( cur << 1|1, mid + 1, r, w);
}

int quiry( int cur, int len, int l, int r){
    if(l == r){
        a[cur].w -= len;
        return a[cur].num;
    }
    int ans;
    int mid = (l + r) >> 1;//优先访问左边点
    if(a[cur << 1].w >= len) ans = quiry( cur << 1, len, l, mid);
    else if(a[cur << 1|1].w >= len) ans = quiry( cur << 1|1, len, mid + 1, r);
    pushup(cur);
    return ans;
}

int main(){
    int h,w,n;
    while(scanf("%d %d %d", &h, &w, &n) != EOF){
        if(h >= n) h = n;//因为h很大,所以不能直接用h来建树
        build(1, 1, h, w);
        while(n--){
            int k;
            scanf("%d", &k);
            if(k > a[1].w)
                printf("-1\n");
            else printf("%d\n", quiry( 1, k, 1, h));
        }
    }
    return 0;
}