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
分析：离线离散化后，按时间插入主席树，最后求下区间个数和即可；
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define Lson L, mid, rt<<1 #define Rson mid+1, R, rt<<1|1 const int maxn=1e5+10; using namespace std; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} int n,m,k,t,a[maxn],b[maxn*2],s[maxn*20],ls[maxn*20],rs[maxn*20],root[maxn*20],sz; inline ll read() { ll x=0;int f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } struct node { int x,y,z; node(){} node(int _x,int _y,int _z):x(_x),y(_y),z(_z){} }op[maxn]; void insert(int l,int r,int x,int &y,int v) { y=++sz; s[y]=s[x]+1; if(l==r)return; ls[y]=ls[x],rs[y]=rs[x]; int mid=l+r>>1; if(v<=mid)insert(l,mid,ls[x],ls[y],v); else insert(mid+1,r,rs[x],rs[y],v); } int query(int l,int r,int L,int R,int x,int y) { if(l==L&&r==R)return s[y]-s[x]; int mid=L+R>>1; if(r<=mid)return query(l,r,L,mid,ls[x],ls[y]); else if(l>mid)return query(l,r,mid+1,R,rs[x],rs[y]); else return query(l,mid,L,mid,ls[x],ls[y])+query(mid+1,r,mid+1,R,rs[x],rs[y]); } int main() { int i,j; scanf("%d",&t); while(t--) { sz=0; scanf("%d%d",&n,&m); rep(i,1,n)a[i]=read(),b[i]=a[i]; rep(i,1,m) { int c,d,e; c=read(),d=read(),e=read(); c++,d++; b[i+n]=e; op[i]=node(c,d,e); } printf("Case %d: ",++k); sort(b+1,b+n+m+1); int num=unique(b+1,b+n+m+1)-b-1; rep(i,1,n) { a[i]=lower_bound(b+1,b+num+1,a[i])-b; insert(1,num,root[i-1],root[i],a[i]); } rep(i,1,m) { int x=op[i].x,y=op[i].y,z=op[i].z; z=lower_bound(b+1,b+num+1,z)-b; printf("%d ",query(1,z,1,num,root[x-1],root[y])); } } //system("Pause"); return 0; }