Billboard——线段树

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题目来源

hduP2795

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题目描述

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

Author
hhanger@zju

Source
HDOJ 2009 Summer Exercise（5）

Recommend
lcy

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题目大意

大学门口有一张 H * W 的广告牌。某一天，这张广告牌空了。需要向广告牌上张贴 1 * wi 大小的广告，广告一定要张贴在最靠上而且最靠左的位置，输出每一张广告张贴的行号，如果无法张贴则输出 -1。

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解题报告

• 本题目是一个线段树模板题目，每一个叶子节点对应一行，记录当前行剩余空间。树的其他节点则记录子节点中剩余空间的最大值。
• 判断是否可以张贴就可以对比根节点的大小与当前广告的宽度。如果根节点的数据小于当前广告的宽度，则说明无法张贴。
• 查询张贴位置的时候优先考虑左侧子树，如果左侧可以张贴则说明位置可以比右侧子树考上，因此无需考虑右侧子树。

此题考察线段树的单点修改单点查询

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代码

#include <iostream>#include <cstdio>#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1using namespace std;int tree[400010 << 2];int h, w, n; // h行, 每个行 w 个空间, n条广告 void PushUP(int rt) {    tree[rt] = max(tree[rt << 1], tree[rt << 1 | 1]);//更新操作为去左右子树中剩余空间最大的}void build(int l, int r, int rt) {    tree[rt] = w;// 建树    if (l == r)        return;    int m = (l + r) >> 1;    build(lson);    build(rson);}int query(int x, int l, int r, int rt) {    if (l == r) {        tree[rt] -= x;        return l;    }    int m = (l + r) >> 1;    int ret = (tree[rt << 1] >= x) ? query(x, lson) : query(x, rson); //选择左右子树张贴广告    PushUP(rt);    return ret;}int main() {    freopen("in.txt", "r", stdin);    while(~scanf("%d%d%d", &h, &w, &n)) {         if (h > n)            h = n;        build(1, h, 1);        int data;        for (int i = 1; i <= n; i++) {            scanf("%d", &data);            if (tree[1] < data)// 判断是否可以张贴                printf("-1\n");            else                printf("%d\n", query(data, 1, h, 1));        }    }     return 0;}