hdu 4417 Super Mario 树状数组||主席树

Super Mario

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

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

 

Source

2012 ACM/ICPC Asia Regional Hangzhou Online、

题意：n个数，m个询问

　　　求区间l，r小于等于h的数目；

思路：主席树求法，单点更新，区间查询；

　　　

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+,M=1e6+,inf=;
const ll INF=1e18+,mod=;
struct Chairmantree
{
    int rt[N*],ls[N*],rs[N*],sum[N*];
    int tot;
    void init()
    {
        tot=;
    }
    void build(int l,int r,int &pos)
    {
        pos=++tot;
        sum[pos]=;
        if(l==r)return;
        int mid=(l+r)>>;
        build(l,mid,ls[pos]);
        build(mid+,r,rs[pos]);
    }
    void update(int p,int c,int pre,int l,int r,int &pos)
    {
        pos=++tot;
        ls[pos]=ls[pre];
        rs[pos]=rs[pre];
        sum[pos]=sum[pre]+c;
        if(l==r)return;
        int mid=(l+r)>>;
        if(p<=mid)
            update(p,c,ls[pre],l,mid,ls[pos]);
        else
            update(p,c,rs[pre],mid+,r,rs[pos]);
    }
    int query(int s,int t,int L,int R,int l,int r)
    {
        if(L<=l&&r<=R)
            return sum[t]-sum[s];
        int mid=(l+r)>>;
        int ans=;
        if(L<=mid)
            ans+=query(ls[s],ls[t],L,R,l,mid);
        if(R>mid)
            ans+=query(rs[s],rs[t],L,R,mid+,r);
        return ans;
    }
};
Chairmantree tree;
int a[N],b[N<<];
int l[N],r[N],x[N];
int getpos(int x,int cnt)
{
    int pos=lower_bound(b+,b+cnt,x)-b;
    return pos;
}
int main()
{
    int T,cas=;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=;i<=n;i++)
        {
            scanf("%d",&a[i]);
            b[i]=a[i];
        }
        //for(int i=1;i<num;i++)
        //    printf("%d ",tree.rt[i]);
        //printf("\n");
        for(int i=;i<=m;i++)
        {
            scanf("%d%d%d",&l[i],&r[i],&x[i]);
            b[i+n]=x[i];
        }
        sort(b+,b++n+m);
        int num=unique(b+,b++n+m)-b;
        tree.init();
        tree.build(,num-,tree.rt[]);
        for(int i=;i<=n;i++)
        {
            int p=getpos(a[i],num);
            tree.update(p,,tree.rt[i-],,num-,tree.rt[i]);
        }
        printf("Case %d:\n",cas++);
        for(int i=;i<=m;i++)
        {
            int p=getpos(x[i],num);
            printf("%d\n",tree.query(tree.rt[l[i]],tree.rt[r[i]+],,p,,num-));
        }
    }
    return ;
}

树状数组求法：

　　按照值的大小排序；

　　标记区间的下标，详见代码；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<" "<<x<<endl;
const int N=1e5+,M=1e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
int n,m;
struct treearray
{
    int tree[N];
    void init()
    {
        memset(tree,,sizeof(tree));
    }
    int lowbit(int x)
    {
        return x&(-x);
    }
    void update(int x,int c)
    {
        while(x<N)
        {
            tree[x]+=c;
            x+=lowbit(x);
        }
    }
    int query(int x)
    {
        int sum=;
        while(x)
        {
            sum+=tree[x];
            x-=lowbit(x);
        }
        return sum;
    }
}treearray;
struct a
{
    int x,pos;
    bool operator <(const a b)const
    {
        return x<b.x;
    }
}a[N];
struct q
{
    int l,r,pos,h;
    bool operator <(const q b)const
    {
        return h<b.h;
    }
}q[N];
int ans[N];
int main()
{
    int T,cas=;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=;i<=n;i++)
            scanf("%d",&a[i].x),a[i].pos=i;
        printf("Case %d:\n",cas++);
        for(int i=;i<=m;i++)
            scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].h),q[i].pos=i;
        sort(q+,q++m);
        sort(a+,a++n);
        treearray.init();
        int st=;
        for(int i=;i<=m;i++)
        {
            //bug(q[i].pos);
            while(st<=n&&a[st].x<=q[i].h)
            {
                //cout<<a[st].pos<<endl;
                treearray.update(a[st].pos,);
                st++;
            }
            ans[q[i].pos]=treearray.query(q[i].r+)-treearray.query(q[i].l);
        }
        for(int i=;i<=m;i++)
            printf("%d\n",ans[i]);
    }
    return ;
}