HDU 4417 离线+树状数组

Super Mario

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6174    Accepted Submission(s): 2675

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：给你n个数 m个查询 l r x  问区间[l,r]中小于等于x的数的个数

 题解：离线+树状数组 将n个数的的位置存入树状数组。

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <map>
 #define ll  __int64
 #define mod 1000000007
 #define dazhi 2147483647
 using namespace  std;
 int t;
 int n,m;
 int tree[];
 int ans[];
 int lowbit(int tt)
 {
     return tt&(-tt);
 }
 void add(int x,int y)
 {
     for(int i=x; i<=n; i+=lowbit(i))
         tree[i]+=y;
 }
 int getsum(int x)
 {
     int ans=;
     for(int i=x; i>; i-=lowbit(i))
         ans+=tree[i];
     return ans;
 }
 struct node
 {
     int w;
     int pos;
 } N[];
 struct xx
 {
     int l,r;
     int x;
     int pos;
 } M[];
 bool cmp1 (struct node aa,struct node bb)
 {
     return aa.w<bb.w;
 }
 bool cmp2 (struct xx aa,struct xx bb)
 {
     return aa.x<bb.x;
 }
 int main()
 {
         scanf("%d",&t);
         for(int i=; i<=t; i++)
         {
             memset(tree,,sizeof(tree));
             scanf("%d %d",&n,&m);
             for(int j=; j<=n; j++)
             {
                 scanf("%d",&N[j].w);
                 N[j].pos=j;
             }
             sort(N+,N++n,cmp1);
             for(int j=; j<=m; j++)
             {
                 scanf("%d %d %d",&M[j].l,&M[j].r,&M[j].x);
                 M[j].l++;
                 M[j].r++;
                 M[j].pos=j;
             }
             sort(M+,M++m,cmp2);
             int exm=;
             for(int j=; j<=m; j++)
             {
                 while(exm<=n&&N[exm].w<=M[j].x)
                 {
                     add(N[exm].pos,);
                     exm++;
                 }
                 ans[M[j].pos]=getsum(M[j].r)-getsum(M[j].l-);
             }
             printf("Case %d:\n",i);
             for(int j=; j<=m; j++)
                 printf("%d\n",ans[j]);
         }
     return ;
 }