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.)
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
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。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <algorithm> 6 #include <stack> 7 #include <queue> 8 #include <cmath> 9 #include <map> 10 #define ll __int64 11 #define mod 1000000007 12 #define dazhi 2147483647 13 using namespace std; 14 int t; 15 int n,m; 16 int tree[100005]; 17 int ans[100005]; 18 int lowbit(int tt) 19 { 20 return tt&(-tt); 21 } 22 void add(int x,int y) 23 { 24 for(int i=x; i<=n; i+=lowbit(i)) 25 tree[i]+=y; 26 } 27 int getsum(int x) 28 { 29 int ans=0; 30 for(int i=x; i>0; i-=lowbit(i)) 31 ans+=tree[i]; 32 return ans; 33 } 34 struct node 35 { 36 int w; 37 int pos; 38 } N[100005]; 39 struct xx 40 { 41 int l,r; 42 int x; 43 int pos; 44 } M[100005]; 45 bool cmp1 (struct node aa,struct node bb) 46 { 47 return aa.w<bb.w; 48 } 49 bool cmp2 (struct xx aa,struct xx bb) 50 { 51 return aa.x<bb.x; 52 } 53 int main() 54 { 55 scanf("%d",&t); 56 for(int i=1; i<=t; i++) 57 { 58 memset(tree,0,sizeof(tree)); 59 scanf("%d %d",&n,&m); 60 for(int j=1; j<=n; j++) 61 { 62 scanf("%d",&N[j].w); 63 N[j].pos=j; 64 } 65 sort(N+1,N+1+n,cmp1); 66 for(int j=1; j<=m; j++) 67 { 68 scanf("%d %d %d",&M[j].l,&M[j].r,&M[j].x); 69 M[j].l++; 70 M[j].r++; 71 M[j].pos=j; 72 } 73 sort(M+1,M+1+m,cmp2); 74 int exm=1; 75 for(int j=1; j<=m; j++) 76 { 77 while(exm<=n&&N[exm].w<=M[j].x) 78 { 79 add(N[exm].pos,1); 80 exm++; 81 } 82 ans[M[j].pos]=getsum(M[j].r)-getsum(M[j].l-1); 83 } 84 printf("Case %d: ",i); 85 for(int j=1; j<=m; j++) 86 printf("%d ",ans[j]); 87 } 88 return 0; 89 }