HDU 2795 Billboard（线段树 区间最大）

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23102    Accepted Submission(s): 9554

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 524333

 

Sample Output

1213-1

 

Author

hhanger@zju

 

题意：

有一块h*w的矩形广告牌，给你n个广告，都是1*wi，都往尽量往左上贴，每个广告输出贴的排数，不能贴了就输出-1。

POINT：

用线段树来记录每排空余的最大值。优先考虑左子树。

线段树的区间代表第几排。l-r代表第l排到r排。

考虑要n只有200000，那即使h特别大，我们也只要n排就行了。这个WA点看代码的注释

#include <iostream>#include <stdio.h>#include <math.h>#include <string.h>using namespace std;const int N = 200088*8;#define LL long longint len[N];int w,h,k;void build(int x,int l,int r){    if(l==r) len[x]=w;    else    {        int mid=(l+r)>>1;        build(2*x,l,mid);        build(2*x+1,mid+1,r);        len[x]=w;    }}void put(int x,int l,int r,int a,int &flag){    if(l==r)    {        flag=1;        len[x]=len[x]-a;        printf("%d\n",l);    }    else    {        int mid=(l+r)>>1;        if(a<=len[x*2]) put(x*2,l,mid,a,flag);        else if(a<=len[x*2+1]) put(x*2+1,mid+1,r,a,flag);        len[x]=max(len[x*2+1],len[x*2]);    }}int main(){    int n;    while(~scanf("%d %d %d",&h,&w,&k))    {        if(h>200000)            n=200000;//改成取h和k的最小值就会WA。        else n=h;        build(1,1,n);        for(int i=1;i<=k;i++)        {            int a;            scanf("%d",&a);            int flag=0;            put(1,1,n,a,flag);            if(!flag) printf("-1\n");        }    }}