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  • HDU 2795

    Billboard
    Time Limit:8000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
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    Status
    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
    <span style="color:#6600cc;">/***************************************************
    
          author   :   Grant Yuan
          time     :   2014.7.27
          algorithm:   线段树
          source   :   HDU 2795
    
    ****************************************************/
    
    #include <iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    #define MAX  200007
    
    using namespace std;
    struct node{
    long long l;
    long long r;
    long long m;
    };
    node tree[3*MAX];
    long long n,h,w,a,ans;
    
    using namespace std;
    
    void build(long long left,long long right,long long root)
    {
        tree[root].l=left;
        tree[root].r=right;
        if(left==right)
        {
            tree[root].m=w;
            return;
        }
        long long mid=(left+right)>>1;
        build(left,mid,root*2);
        build(mid+1,right,root*2+1);
        tree[root].m=max(tree[root*2].m,tree[root*2+1].m);
    }
    
    void update(int mm,int root)
    {
        if(mm>tree[root].m)
            return;
        if(tree[root].l==tree[root].r){
            ans=tree[root].l;
            tree[root].m-=mm;
            return;
        }
        if(tree[root*2].m>=mm)
            update(mm,root*2);
        else if(tree[root*2+1].m>=mm)
            update(mm,root*2+1);
        tree[root].m=max(tree[root*2].m,tree[root*2+1].m);
    }
    
    int main()
    {
       while(~scanf("%lld%lld%lld",&h,&w,&n)){
            if(h>n) h=n;
            build(1,h,1);
            for(int i=1;i<=n;i++)
            {
                ans=-1;
                scanf("%lld",&a);
                update(a,1);
                printf("%lld
    ",ans);
            }
       }
        return 0;
    }
    </span>

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  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254479.html
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