Billboard

Time Limit: 20000/8000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32864 Accepted Submission(s): 13090

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

题意：多组输入。先给三个整数，h，w，n。分别代表墙的高度，宽度，和海报的数量。接下来n行每行一个数代表这张海报的宽度(高度默认为1）

思路：线段树维护墙的高度区间上的最大值。

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define ll long long
#define inf 0x3f
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
const int maxn = 200000 + 5;
using namespace std;
int h,n,w;
struct Segment
{
    int l,r;
    int sum;
    #define l(x) tree[x].l
    #define r(x) tree[x].r
    #define sum(x) tree[x].sum
}tree[maxn * 4];
void build(int q,int l,int r)
{
    l(q) = l,r(q) = r;
    if(l == r)
    {
        sum(q) = w;
        return ;
    }
    int mid = (l + r) / 2;
    build(q*2,l,mid);
    build(q*2+1,mid+1,r);
    sum(q) = max(sum(q*2),sum(q*2+1));
}
void clearr()
{
    memset( tree,0 ,sizeof(tree));
}
void ask(int q,int len)
{
    if(l(q) == r(q))
    {
        sum(q) -= len;
        //cout<<q<<endl;
        printf("%d\n",l(q) );
        return ;
    }
    if(sum(q*2) >= len) ask(q*2,len);
    else if(sum(q*2+1) >= len) ask(q*2+1,len);
    else printf("-1\n");//cout<<q<<endl;}
    sum(q) = max(sum(q*2),sum(q*2+1));
}
int main()
{

    //freopen("C:\\ACM\\input.txt","r",stdin);
    while(scanf("%d%d%d",&h,&w,&n) != EOF)
    {
        clearr();
        if(h>n)
            h=n;
        build(1,1,h);
        for(int i = 1;i <= n; ++i)
        {
            int len;
            scanf("%d",&len);
            if(sum(1) >= len )
                ask(1,len);
            else printf("-1\n");
        }

    }
}

原文地址：https://www.cnblogs.com/jrfr/p/11531877.html

HDU 2795(思维+线段树）

Billboard