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HDU4417-Super Mario

2024-07-09 04:08:25   220人阅读

Super Mario

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2183    Accepted Submission(s): 1061


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题目意思就是一条线段上有n个点，每个点有个高度值，然后有m组查询，问你a到b区间里有多少个数是小于等于h的这道题的解法为用二分+划分树二分的是第K大数，与高度值进行比较好久没1A了。。。。QAQ
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
#define MD(x,y) (((x)+(y))>>1)
const int maxn = 100000+10;
int lftnum[20][maxn];
int num[maxn];
int seg[20][maxn];
int n,m;
void build(int L,int R,int dep){
    if(L==R) return;
    int mid = MD(L,R),key = num[mid],lcnt = mid-L+1;
    for(int i = L; i <= R; i++){
        if(seg[dep][i] < key)
            lcnt--;
    }
    int lp = L,rp = mid+1;
    for(int i = L; i <= R; i++){
        if(L==i){
            lftnum[dep][i] = 0;
        }else{
            lftnum[dep][i] = lftnum[dep][i-1];
        }
        if(seg[dep][i] < key){
            lftnum[dep][i]++;
            seg[dep+1][lp++] = seg[dep][i];
        }
        else if(seg[dep][i] > key){
            seg[dep+1][rp++] =seg[dep][i];
        }
        else{
            if(lcnt>0){
                lcnt--;
                lftnum[dep][i]++;
                seg[dep+1][lp++] = seg[dep][i];
            }else{
                seg[dep+1][rp++] = seg[dep][i];
            }
        }
    }
    build(L,mid,dep+1);
    build(mid+1,R,dep+1);
}
int query(int L,int R,int ll,int rr,int dep,int k){
    if(L==R)
        return seg[dep][L];
    int a,aa,b,bb;
    int mid = MD(ll,rr);
    if(L==ll){
        a  = 0;
        aa = lftnum[dep][R] - a;
    }else{
        a = lftnum[dep][L-1];
        aa = lftnum[dep][R] - a;
    }
    if(aa >= k){
        L = ll+a;
        R = ll+a+aa-1;
        return query(L,R,ll,mid,dep+1,k);
    }else{
        b = L-ll-a;
        bb = R-L+1-aa;
        L = mid+1+b;
        R = mid+b+bb;
        return query(L,R,mid+1,rr,dep+1,k-aa);
    }
}
int main(){

    int ncase,T=1;
    cin >> ncase;
    while(ncase--){
        scanf("%d%d",&n,&m);
        for(int i = 1; i <= n; i++){
            scanf("%d",&num[i]);
            seg[0][i] = num[i];
        }
        sort(num+1,num+1+n);
        build(1,n,0);
        printf("Case %d:\n",T++);
        while(m--){
            int a,b,h;
            scanf("%d%d%d",&a,&b,&h);
            ++a,++b;
            int l = 1,r = b-a+1;
            while(l <= r){
                int mid = MD(l,r);
                if(query(a,b,1,n,0,mid) > h){
                    r = mid-1;
                }else{
                    l = mid+1;
                }
            }
            printf("%d\n",r);
        }
    }
    return 0;
}



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