Super Mario

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6077    Accepted Submission(s): 2645

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

【分析】给你一系列数，求给定区间内不大于h的数的个数。直接二分枚举位置然后划分树查找就行了。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <time.h>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define lson(x) ((x<<1))
#define rson(x) ((x<<1)+1)
using namespace std;
typedef long long ll;
const int N=1e5+50;
const int M=N*N+10;
long long ans;
struct P_Tree {
    int n;
    int tree[20][N];
    int sorted[N];
    int toleft[20][N];
    ll sum[N];
    ll lsum[20][N];

    void init(int len) {
        n=len;
        sum[0]=0;
        for(int i=0; i<20; i++)tree[i][0]=toleft[i][0]=lsum[i][0]=0;
        for(int i=1; i<=n; i++) {
            scanf("%d",&sorted[i]);
            tree[0][i]=sorted[i];
            sum[i]=sum[i-1]+sorted[i];
        }
        sort(sorted+1,sorted+n+1);
        build(1,n,0);
    }
    void build(int l,int r,int dep) {
        if(l==r)return;
        int mid=(l+r)>>1;
        int same=mid-l+1;
        for(int i=l; i<=r; i++) {
            if(tree[dep][i]<sorted[mid])
                same--;
        }
        int lpos = l;
        int rpos = mid+1;
        for(int i = l; i <= r; i++) {
            if(tree[dep][i] < sorted[mid]) {
                tree[dep+1][lpos++] = tree[dep][i];
                lsum[dep][i] = lsum[dep][i-1] + tree[dep][i];
            } else if(tree[dep][i] == sorted[mid] && same > 0) {
                tree[dep+1][lpos++] = tree[dep][i];
                lsum[dep][i] = lsum[dep][i-1] + tree[dep][i];
                same --;
            } else {
                tree[dep+1][rpos++] = tree[dep][i];
                lsum[dep][i] = lsum[dep][i-1];
            }
            toleft[dep][i] = toleft[dep][l-1] + lpos -l;
        }
        build(l,mid,dep+1);
        build(mid+1,r,dep+1);
    }

    int query(int L,int R,int l,int r,int dep,int k) {
        if(l==r)return tree[dep][l];
        int mid=(L+R)>>1;
        int cnt=toleft[dep][r]-toleft[dep][l-1];
        int s=toleft[dep][l-1]-toleft[dep][L-1];
        if(cnt>=k) {
            int newl=L+toleft[dep][l-1]-toleft[dep][L-1];
            int newr=newl+cnt-1;
            return query(L,mid,newl,newr,dep+1,k);//×￠òa
        } else {
            ans+=(ll)(lsum[dep][r]-lsum[dep][l-1]);
            //printf("l:  %d   r:   %d  ans:  %lld   %lld  %lld\n",l,r,ans,lsum[dep][r],lsum[dep][l-1]);
            int newr=r+toleft[dep][R]-toleft[dep][r];
            int newl=newr-(r-l-cnt);
            return query(mid+1,R,newl,newr,dep+1,k-cnt);//×￠òa
        }
    }
} tre;
int main() {
    int iCase=0;
    int n,m,k,T;
    int l,r,h;
    scanf("%d\n",&T);
    while(T--) {
        scanf("%d%d",&n,&m);
        tre.init(n);
        printf("Case %d:\n",++iCase);
        while(m--) {
            int Ans=0;
            ans=0;
            scanf("%d%d%d",&l,&r,&h);
            l++;
            r++;
            int ll=1,rr=(r-l)+1;
            while(rr-ll>=0){
                k=(rr+ll)/2;
                int res=tre.query(1,n,l,r,0,k);
                if(res>h)rr=k-1;
                else {
                    Ans=k;
                    ll=k+1;
                }
                //printf("l: %d r: %d k: %d ans: %d\n",l,r,k,Ans);system("pause");
            }
            printf("%d\n",Ans);
        }
    }
    return 0;
}