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  • hdu4177:Super Mario

    主席树+离散化。给一段区间。多次询问[l,r]中有多少个数小于k。啊主席树用指针版写出来优美多了QAQ。。。

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define REP(i,s,t) for(int i=s;i<=t;i++)
    #define dwn(i,s,t) for(int i=s;i>=t;i--)
    #define clr(x,c) memset(x,c,sizeof(x))
    int read(){
    	int x=0;char c=getchar();bool f=true;
    	while(!isdigit(c)){
    		if(c=='-') f=false;c=getchar();
    	}
    	while(isdigit(c)) x=x*10+c-'0',c=getchar();
    	return f?x:-x;
    } 
    const int nmax=100005;
    const int inf=0x7f7f7f7f;
    struct node{
    	node *l,*r;int sum;
    };
    node *root[nmax],nodes[nmax*20],*pt;
    int a[nmax],b[nmax];
    node* build(int l,int r){
    	node* op=pt++;op->sum=0;
    	if(l==r) return op;
    	int mid=(l+r)>>1;
    	op->l=build(l,mid);op->r=build(mid+1,r);
    	return op;
    }
    node* update(int p,int add,node* t,int l,int r){
    	node* op=pt++;op->sum=t->sum+add;
    	if(l==r) return op;
    	int mid=(l+r)>>1;
    	if(p<=mid) op->r=t->r,op->l=update(p,add,t->l,l,mid);
    	else op->l=t->l,op->r=update(p,add,t->r,mid+1,r);
    	return op;
    }
    int query(node* t,node* s,int x,int l,int r){
    	if(l==r) return t->sum-s->sum;
    	int mid=(l+r)>>1;
    	if(x<=mid) return  query(t->l,s->l,x,l,mid);
    	else return query(t->r,s->r,x,mid+1,r)+t->l->sum-s->l->sum;
    }
    int main(){
    	int cas=read();
    	REP(i,1,cas){
    		pt=nodes;
    		int n=read(),m=read();
    		REP(j,1,n) a[j]=b[j]=read();
    		sort(b+1,b+n+1);
    		int N=unique(b+1,b+n+1)-b-1;
    		N++;b[N]=inf;
    		
    		root[0]=build(1,N);
    		REP(j,1,n){
    			int k=lower_bound(b+1,b+N+1,a[j])-b;
    			root[j]=update(k,1,root[j-1],1,N);
    		}
    
    		printf("Case %d:
    ",i);
    		REP(j,1,m){
    			int l=read(),r=read(),v=read();l++,r++;
    			int k=upper_bound(b+1,b+1+N,v)-b-1;
    			int ans=0;
    			if(k>0) ans=query(root[r],root[l-1],k,1,N);
    			printf("%d
    ",ans);
    		}
    	}
    	return 0;
    }
    

      

    Super Mario

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4662 Accepted Submission(s): 2148


    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
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  • 原文地址:https://www.cnblogs.com/fighting-to-the-end/p/5716385.html
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