HDU-4417-Super Mario（线段树+离线）

Super Mario

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8331    Accepted Submission(s): 3523

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次查询，每次求在l到r区间内小于等于H的书有多少个；
思路：这题很显然需要离线处理，每个数从小到大排序，每个询问按H的大小从小到大排序，对于每次询问，把小于等于H的数更新到线段树中，然后区间求和即可；
AC代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
struct node
{
    int x,pos;
    node(){};
    node(int tx,int tpos){x=tx;pos=tpos;}
}arr[maxn];
bool cmp1(node a,node b)
{
    if(a.x==b.x) return a.pos<b.pos;
    return a.x<b.x;
}
struct ques
{
    int l,r,k,id;
}Q[maxn];
bool cmp(ques a,ques b)
{
    if(a.k==b.k){
        if(a.l==b.l) return a.r<b.r;
        else return a.l<b.l;
    }
    return a.k<b.k;
}

struct NODE
{
    int l,r;
    int sum;
}tree[maxn*4];
void build(int root,int l,int r)
{
    tree[root].l=l;
    tree[root].r=r;
    tree[root].sum=0;
    if(l==r) return;
    int mid=(l+r)>>1;
    build(root<<1,l,mid);
    build(root<<1|1,mid+1,r);
}
void pushup(int root)
{
    tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum;
}
void update(int root,int pos)
{
    int ll=tree[root].l;
    int rr=tree[root].r;
    if(ll==rr){
        tree[root].sum++;
        return;
    }
    int mid=(ll+rr)>>1;
    if(pos<=mid) update(root<<1,pos);
    else update(root<<1|1,pos);
    pushup(root);
}
int query(int root,int l,int r)
{
    int ll=tree[root].l;
    int rr=tree[root].r;
    if(l<=ll&&rr<=r)
        return tree[root].sum;
    int ans=0;
    int mid=(ll+rr)>>1;
    if(l<=mid) ans+=query(root<<1,l,r);
    if(r>mid) ans+=query(root<<1|1,l,r);
    return ans;
}
int n,m,ans[maxn];
int main()
{
    int T;
    scanf("%d",&T);
    int cas=1;
    while(T--)
    {
        scanf("%d%d",&n,&m);
        int x;
        for(int i=1;i<=n;i++){
            scanf("%d",&x);
            arr[i]=node(x,i);
        }
        int l,r,k;
        for(int i=1;i<=m;i++){
            scanf("%d%d%d",&Q[i].l,&Q[i].r,&Q[i].k);
            Q[i].l++;Q[i].r++;
            Q[i].id=i;
        }
        sort(Q+1,Q+1+m,cmp);
        sort(arr+1,arr+1+n,cmp1);
        build(1,1,n);
        int cnt=1;
        for(int i=1;i<=m;i++){
            while(cnt<=n&&arr[cnt].x<=Q[i].k){
                update(1,arr[cnt].pos);
                cnt++;
            }
            ans[Q[i].id]=query(1,Q[i].l,Q[i].r);
        }
        printf("Case %d:\n",cas++);
        for(int i=1;i<=m;i++){
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}