Billboard HDU - 2795---线段树--单点修改--区间最大值--优先遍历左树

Dear 丶 2023-07-24 08:40 2阅读 0赞

## 题目描述 ##

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

## 简单说下这题的写法 ##

题目的意思是问高度为1宽度为w的广告能放在告示栏的哪个地方，若不能放则输出-1，放的规则是优先上方，优先左边。  
这道题目怎么和线段树联系在一起呢？  
想想看高度为1的广告和高度为h的告示栏，我们是不是可以想象成h个区间？每个区间对应的初值为w，每次放入一个广告时，减去相应的w。我们只需要优先遍历左树即可。  
要注意的是这颗线段树存储的不是区间和而是区间的最大值，我们通过判断广告的宽度和区间最大值的大小关系来进行树的遍历。  
代码如下：

#include<cstdio>
    #include<algorithm>
    #include<cstring>
    using namespace std;
    #define ll long long
    inline ll maxx(ll x,ll y)
    {
        
        return x>y?x:y;
    }
    typedef struct
    {
        
        ll w,maxn,tag,l,r;
    } NODE;
    NODE tree[800005];
    ll h,w,n,cur=1,ans;
    void build_tree(ll x,ll y,ll k)
    {
        
        tree[k].l=x;
        tree[k].r=y;
        if(tree[k].l==tree[k].r)
        {
        
            tree[k].tag=cur++;
            tree[k].w=w;
            tree[k].maxn=w;
            return ;
        }
        ll mid=x+(y-x)/2;
        build_tree(x,mid,k<<1);
        build_tree(mid+1,y,k<<1|1);
        tree[k].maxn=maxx(tree[k<<1].maxn,tree[k<<1|1].maxn);
    }
    void chang_for_one(ll ambition,ll k)
    {
        
        if(ambition>tree[k].maxn)
        {
        
            ans=-1;
            return ;
        }
        if(tree[k].l==tree[k].r)
        {
        
            tree[k].maxn-=ambition;
            tree[k].w-=ambition;
            ans=tree[k].tag;
            return ;
        }
        if(tree[k<<1].maxn>=ambition)
        {
        
            chang_for_one(ambition,k<<1);
        }
        else if(tree[k<<1|1].maxn>=ambition)
        {
        
            chang_for_one(ambition,k<<1|1);
        }
        tree[k].maxn=maxx(tree[k<<1].maxn,tree[k<<1|1].maxn);
    }
    int main()
    {
        
        while(scanf("%lld %lld %lld",&h,&w,&n)!=EOF)
        {
        
            cur=1;
            if(h<n)
            build_tree(1,h,1);
            else
                build_tree(1,n,1);
            while(n--)
            {
        
                ll x;
                scanf("%lld",&x);
                chang_for_one(x,1);
                printf("%lld\n",ans);
            }
        }
        return 0;
    }