線段樹練習[單點更新] HDU 2795 Billboard

題目大意：有一個h*w的公告榜，可以依次在上面添加資訊。每個資訊的長度為x，高為1. 

優先在最上面加入，如果空間足夠的話，然後優先放在最左面。統計每條公告最終的位置，即它所在的行數。

這裡是線段樹來儲存 當前區間(i,j)的所有位置，剩餘的最大空間。 初始化即為w，公告榜的寬。

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

 

#include <iostream>#include <stdio.h>using namespace std;int tree[1000000]; // 記錄當前區間所有位置，剩餘的最大空間int h,w,n,t;int max(int a,int b){    return a > b ? a : b;}void build(int l,int r,int k) {    tree[k] = w;    if (l == r) return ;    int m = (l + r) >> 1;    build(l,m,k*2);    build(m+1,r,k*2+1);}void update(int t,int l, int r, int k){    if(t > tree[k]){ // 如果當前無法加入，就直接返回        printf("-1\n");        return;    }    if(l == r){ //說明找到了合適位置可加入。        printf("%d\n", l);        tree[k] -= t;        return;    }    int m = (l+r)/2;    if(t <= tree[k*2])        update(t, l, m, k*2);  //優先加入左子樹，即上面    else if(t <= tree[k*2+1])        update(t, m+1, r, k*2+1);    tree[k] = max(tree[k*2], tree[2*k+1]);}int main() {//freopen("in.txt", "r" ,stdin);    while(scanf("%d %d %d", &h, &w, &n) != EOF){        if(h > n)            h = n;        build(1,h,1);        while(n--){ //更新           scanf("%d", &t);            update(t, 1, h ,1);        }    }    return 0;}