HDU-4417-Super Mario(划分树+二分)

最新推荐文章于 2020-10-19 15:36:44 发布

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最新推荐文章于 2020-10-19 15:36:44 发布

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分类专栏：划分树

本文链接：https://blog.youkuaiyun.com/faithdmc/article/details/37668709

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本文探讨了在给定的最大跳跃高度下，如何计算马里奥在指定路径区间内可以击中的砖块数量。通过使用划分树和二分查找的方法，有效地解决了这一问题。案例分析和输入输出示例提供了具体的实现步骤和结果。

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Problem Description

Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.

Input

The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)

Output

For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.

Sample Input

  
   1
10 10
0 5 2 7 5 4 3 8 7 7 
2 8 6
3 5 0
1 3 1
1 9 4
0 1 0
3 5 5
5 5 1
4 6 3
1 5 7
5 7 3

Sample Output

Source

2012 ACM/ICPC Asia Regional Hangzhou Online

思路：求区间内不大于给定数的最大数在该区间的第K大。求第K大可以用划分树，然后就二分一下K就好了。

#include <cstdio>
#include <algorithm>
using namespace std;

int sum[20][100005],node[20][100005],sorted[100005];

void build(int c,int s,int e)
{
    int mid,lp,rp,lm,i;

    mid=(s+e)>>1;
    lm=0;
    lp=s;
    rp=mid+1;

    for(i=s;i<=mid;i++) if(sorted[i]==sorted[mid]) lm++;

    for(i=s;i<=e;i++)
    {
        if(s==i) sum[c][i]=0;
        else sum[c][i]=sum[c][i-1];

        if(node[c][i]==sorted[mid])
        {
            if(lm)
            {
                lm--;
                sum[c][i]++;
                node[c+1][lp++]=node[c][i];
            }
            else node[c+1][rp++]=node[c][i];
        }
        else if(node[c][i]<sorted[mid])
        {
            sum[c][i]++;
            node[c+1][lp++]=node[c][i];
        }
        else node[c+1][rp++]=node[c][i];
    }

    if(s!=e)
    {
        build(c+1,s,mid);
        build(c+1,mid+1,e);
    }
}

int query(int c,int s,int e,int l,int r,int k)
{
    if(s==e) return node[c][s];
    else
    {
        int ls,rs,mid;

        if(s==l)
        {
            ls=0;
            rs=sum[c][r];
        }
        else
        {
            ls=sum[c][l-1];
            rs=sum[c][r]-ls;
        }

        mid=(s+e)>>1;

        if(rs>=k) return query(c+1,s,mid,s+ls,s+ls+rs-1,k);
        else return query(c+1,mid+1,e,mid-s+1-ls+l,mid-s+1-ls-rs+r,k-rs);
    }
}

int main()
{
    int T,n,m,i,a,b,c,l,r,mid,ans,casenum;

    scanf("%d",&T);

    for(casenum=1;casenum<=T;casenum++)
    {
        scanf("%d%d",&n,&m);

        for(i=1;i<=n;i++)
        {
            scanf("%d",&sorted[i]);

            node[0][i]=sorted[i];
        }

        sort(sorted+1,sorted+n+1);

        build(0,1,n);

        printf("Case %d:\n",casenum);

        while(m--)
        {
            scanf("%d%d%d",&a,&b,&c);

            if(query(0,1,n,a+1,b+1,1)>c)//特判
            {
                printf("0\n");

                continue;
            }

            l=1,r=b-a+1;

            while(l<=r)//二分
            {
                mid=(l+r)>>1;

                if(query(0,1,n,a+1,b+1,mid)<=c)
                {
                    ans=mid;
                    l=mid+1;
                }
                else
                {
                    r=mid-1;
                }
            }

            printf("%d\n",ans);
        }
    }
}