「super mario」Super Mario

super mario

Time limit: 2000/1000 MS (java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 8448 Accepted Submission(s): 3568

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

Case 1: 4 0 0 3 1 2 0 1 5 1

这题用划分树做，再用二分法查找。

#include<bits/stdc++.h>
using namespace std;
int n,m,tree[20][100001],toleft[20][100001],t,sortnum[100001];
void dfs(int l,int r,int depth)
{
    if(l != r)
    {
        int same = 0,mid = (l + r) / 2 + 1;
        for(int i = l;i < mid;++i)
            if(sortnum[i] == sortnum[mid])
                ++same;
        int temp = sortnum[mid],num = 0,lpos = l,rpos = mid;
        for(int i = l;i <= r;++i)
            if(tree[depth][i] < temp)
            {
                toleft[depth][i] = ++num;
                tree[depth + 1][lpos++] = tree[depth][i];
            }
            else if(tree[depth][i] > temp ||(tree[depth][i] == temp && !same))
            {
                toleft[depth][i] = num;
                tree[depth + 1][rpos++] = tree[depth][i];
            }
            else
            {
                toleft[depth][i] = ++num;
                tree[depth + 1][lpos++] = tree[depth][i];
                --same;
            }
         dfs(l,mid - 1,depth + 1);
         dfs(mid,r,depth + 1);
    }
}
int findk(int L,int R,int l,int r,int k,int depth)
{
    if(l == r)
        return tree[depth][l];
    int temp,temp1 = 0,mid = (L + R) / 2 + 1;
    if(l > L)
        temp1 = toleft[depth][l - 1];
    temp = toleft[depth][r] - temp1;
    if(temp < k)
    {
        int temp2 = mid + l - L - temp1;
        int temp3 = temp2 + r - l - temp;
        return findk(mid,R,temp2,temp3,k - temp,depth + 1);
    }
    return findk(L,mid - 1,L + temp1,L + toleft[depth][r] - 1,k,depth + 1);
}
int findh(int l,int r,int h,int depth)
{
    int ans = 0;
    int left = l,right = r;
    while(left <= right)
    {
        int mid = (right + left) / 2;
        int temp = findk(0,n - 1,l,r,mid - l + 1,0);
        if(temp <= h)
        {
            ans += (mid - left + 1);
            left = mid + 1;
        }
        else
            right = mid - 1;
    }

    return ans;
}
int main()
{
    scanf("%d",&t);
    int casei = 0;
    while(t)
    {
        ++casei;
        scanf("%d",&n);
        scanf("%d",&m);
        for(int i = 0;i < n;++i)
        {
            scanf("%d",&tree[0][i]);
            sortnum[i] = tree[0][i];
        }
        sort(sortnum,sortnum + n);
        dfs(0,n - 1,0);
        printf("Case %d:\n",casei);
        while(m)
        {
            int l,r,k;
            scanf("%d",&l);
            scanf("%d",&r);
            scanf("%d",&k);
            int ans = findh(l,r,k,0);
            printf("%d\n",ans);
            --m;
        }
        --t;
    }
    return 0;
}

Super Mario

super mario

相关阅读

栏目导航

推荐阅读

热门阅读