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

題目意思就是一條線段上有n個點，每個點有個高度值，然後有m組查詢，問你a到b區間裡有多少個數是小於等於h的
這道題的解法為用二分+劃分樹
二分的是第K大數，與高度值進行比較

好久沒1A了。。。。QAQ

#include 
#include 
#include 
#include 
#include 
#include 
#include 
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;
}