hdu 2795

http://acm.hdu.edu.cn/showproblem.php?pid=2795

Billboard

Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30369    Accepted Submission(s): 12230

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

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn=200000+2;

struct segtree
{
    int l,r,maxx;//用树存每一段高度的地方的最大可以存的值，从而判断输入的数可以存在哪里；
};
segtree tree[maxn*3+2];

int T,H,W,N;
int arr[maxn];
int sum,flag;

void build(int n,int l,int r)
{
    tree[n].l=l; tree[n].r=r; tree[n].maxx=W;
    if(l==r)
    {
        return;
    }
    int mid=(tree[n].l+tree[n].r)/2;
    build(n*2,l,mid);
    build(n*2+1,mid+1,r);
}

void updata(int n,int a)
{
    if(flag==1)
    {
        return;
    }
    if(tree[n].l==tree[n].r)
    {
    //    printf("!!
");
    //    printf("!%d %d!
",a,tree[n].maxx);
        if(tree[n].maxx>=a)
        {
            
            flag=1;
            tree[n].maxx-=a; sum=tree[n].l;
        //    printf("%d
",tree[n].maxx);
        }
        return;
    }
    else
    {
        int mid=(tree[n].l+tree[n].r)/2;
        if(tree[n*2].maxx>=a)
        {
            updata(n*2,a);
        }
        else 
        {
            updata(n*2+1,a);
        }
    }
    tree[n].maxx=max(tree[n*2].maxx,tree[n*2+1].maxx);
}

int main()
{
    int a;
//    scanf("%d",&T);
    while(~scanf("%d%d%d",&H,&W,&N))
    {
        if(H>N)当H>N时，比N大的地方没有意义，因为一定不会存在那里；所以题目中H的范围只是一个误导人的陷阱；
        {
            H=N;//这是必须的，没有就会溢出；因为H太大了；
        } 
        build(1,1,H);
        for(int i=1;i<=N;i++)
        {
            sum=-1; flag=0;
            scanf("%d",&a);
            updata(1,a);
            arr[i]=sum;
        }
        for(int i=1;i<=N;i++)
        {
            printf("%d
",arr[i]);
        }
    }
    return 0;
} 

 

#include<cstdio>
#include<algorithm> 
using namespace std;
const int maxn=2e5+5;
struct {
    int l,r,rest;
}tree[maxn*3+3];
int h,w,n,wi,ans;
void build(int s,int l,int r){
    tree[s].l=l;
    tree[s].r=r;
    tree[s].rest=w;
    if(l==r){
        return;
    }
    int m=(l+r)/2;
    build(s*2,l,m);
    build(s*2+1,m+1,r);
}
void updata(int s,int l,int r,int wi){
    if(tree[s].rest<wi){
        return;
    }else if(l==r){
        if(tree[s].rest>=wi){
            tree[s].rest-=wi;
            ans=l;
        }
        return;
    }else{
        int m=(l+r)/2;
        if(tree[s*2].rest>=wi){
            updata(s*2,l,m,wi);
        }else if(tree[s*2+1].rest>=wi&&tree[s*2].rest<wi){
            updata(s*2+1,m+1,r,wi);
        }
        tree[s].rest=max(tree[s*2].rest,tree[s*2+1].rest);
    }
}
int main(){
    while(~scanf("%d%d%d",&h,&w,&n)){
        if(h>n){
            h=n;
        }
        build(1,1,h);
        while(n--){
            scanf("%d",&wi);
            ans=-1;
            updata(1,1,h,wi);
            printf("%d
",ans);
        }
    }
    return 0;
}