摘要：
Super Mario SuperMarioTimeLimit:2000/1000MS（Java/其他）MemoryLimit:32768/32768K（Java/其它）提交总数：6208接受的提交数：2687问题描述Mario是世界著名的水管工。他的“魁梧”身材
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6208 Accepted Submission(s): 2687
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
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 #define N 100005 14 using namespace std; 15 struct chairmantree 16 { 17 int tot; 18 int rt[20*N],ls[20*N],rs[20*N],sum[20*N]; 19 void init() 20 { 21 tot=0; 22 } 23 void buildtree(int l,int r,int &pos) 24 { 25 pos=++tot; 26 sum[pos]=0; 27 if(l==r) return ; 28 int mid=(l+r)>>1; 29 buildtree(l,mid,ls[pos]); 30 buildtree(mid+1,r,rs[pos]); 31 } 32 void update(int p,int c,int pre,int l,int r,int &pos) 33 { 34 pos=++tot; 35 ls[pos]=ls[pre]; 36 rs[pos]=rs[pre]; 37 sum[pos]=sum[pre]+c; 38 if(l==r) return ; 39 int mid=(l+r)>>1; 40 if(p<=mid) 41 update(p,c,ls[pre],l,mid,ls[pos]); 42 else 43 update(p,c,rs[pre],mid+1,r,rs[pos]); 44 } 45 int query(int L,int R,int s,int t,int l,int r) 46 { 47 if(s<=l&&r<=t) 48 return sum[R]-sum[L]; 49 int mid=(l+r)>>1; 50 int ans=0; 51 if(s<=mid) 52 ans+=query(ls[L],ls[R],s,t,l,mid); 53 if(t>mid) 54 ans+=query(rs[L],rs[R],s,t,mid+1,r); 55 return ans; 56 } 57 }tree; 58 59 int t; 60 int n,m; 61 int a[N]; 62 int b[2*N]; 63 int l[N],r[N],x[N]; 64 int getpos(int x,int cnt) 65 { 66 int pos=lower_bound(b+1,b+1+cnt,x)-b; 67 return pos; 68 } 69 int main() 70 { 71 scanf("%d",&t); 72 int T=t; 73 while(t--) 74 { 75 scanf("%d %d",&n,&m); 76 for(int i=1;i<=n;i++){ 77 scanf("%d",&a[i]); 78 b[i]=a[i]; 79 } 80 for(int i=1;i<=m;i++){ 81 scanf("%d %d %d",&l[i],&r[i],&x[i]); 82 b[i+n]=x[i]; 83 } 84 sort(b+1,b+1+n+m); 85 int num=unique(b+1,b+1+n+m)-b; 86 tree.init(); 87 tree.buildtree(1,num-1,tree.rt[0]); 88 for(int i=1;i<=n;i++) 89 { 90 int pos=getpos(a[i],num-1); 91 tree.update(pos,1,tree.rt[i-1],1,num-1,tree.rt[i]); 92 } 93 printf("Case %d: ",T-t); 94 for(int i=1;i<=m;i++) 95 { 96 int pos=getpos(x[i],num-1); 97 printf("%d ",tree.query(tree.rt[l[i]],tree.rt[r[i]+1],1,pos,1,num-1)); 98 } 99 } 100 return 0; 101 }