POJ 2104 K-th Number

题意:给定一个数组,输入i, j, k,也就是询问i到j之间第k大的数

思路:暴力,主席树(模板题);


1、

暴力就是对每一个数标好顺序,然后排序(这时候预处理完成);

对于每个查询,暴力的扫一遍,遇到序号在i,j之间的数就k--,k=0是,就是所找的数

/*暴力法
Time: 7563MS
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100005;
struct node{
	int val,id;
}nod[N];
bool cmp(node a, node b){
	return a.val < b.val;
}
int main(){
	int n, m, l, r, k;
	while(~scanf("%d%d",&n,&m)){
		for(int i = 1; i <= n; i++){
			scanf("%d", &nod[i].val);
			nod[i].id = i;
		}
		sort(nod+1, nod+n+1, cmp);
		for(int i = 0; i < m; i++){
			scanf("%d%d%d", &l, &r, &k);
			for(int j = 1; j <= n; j++){
				if(nod[j].id >= l && nod[j].id <= r) k--;
				if(k == 0){
					printf("%d\n", nod[j].val);
					break;
				}
			}
		}
	}
	return 0;
}

2、

用a[N]存储原来的数字,a2[size]存储排好序的数字;

主席树,T[i]都维护着size大小的线段树,而该线段树维护的信息为:在a[i] 到a[n]数字集合中,size种数字分别出现的次数。

我是看着这个理解的

http://blog.csdn.net/sprintfwater/article/details/9162041

1422MS

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 100005;
const int M = N * 30;
int n,q,m,tot;
int a[N],t[N];
int T[M],lson[M],rson[M],c[M];
void Init_hash(){
	for(int i = 1; i <= n ;i++)
		t[i] = a[i];
	sort(t+1, t+n+1);
	m=unique(t+1,t+1+n)-t-1;
}
int build(int l, int r) {
	int root = tot++;
	c[root] = 0;
    if(l != r){
		int mid = (l+r)>>1;
		lson[root] = build(l, mid);
		rson[root] = build(mid+1, r);
	}
	return root;
}
int hash(int x){
	return lower_bound(t+1, t+1+m, x)-t;
}
int update(int root, int pos, int val){
	int newroot = tot++, tmp = newroot;
	c[newroot] = c[root] + val;
	int l = 1, r = m;
	while(l < r){
		int mid = (l+r)>>1;
		if(pos <= mid){
			lson[newroot] = tot++;
			rson[newroot] = rson[root];
			newroot = lson[newroot];
			root = lson[root];
			r = mid;
		}
		else {
			rson[newroot] = tot++;
			lson[newroot] = lson[root];
			newroot = rson[newroot];
			root = rson[root];
			l = mid+1;
		}
		c[newroot] = c[root] + val;
	}
	return tmp;
}
int query(int lr, int rr, int k){
	int l = 1, r = m;
	while( l < r){
		int mid = (l+r)>>1;
		if(c[lson[lr]] - c[lson[rr]] >= k){
			r = mid;
			lr = lson[lr];
			rr = lson[rr];
		}
		else {
			l = mid + 1;
			k -= c[lson[lr]] - c[lson[rr]];
			lr = rson[lr];
            rr = rson[rr];
		}

	}
	return l;
}
int main(){
	while(~scanf("%d%d",&n,&q)){
		tot=0;
		for(int i = 1; i <= n; i++)
			scanf("%d",a+i);
		Init_hash();
		T[n+1] = build(1, m);
		for(int i = n; i ; i--){
			int pos = hash(a[i]);
            T[i] = update(T[i+1],pos,1);
		}
        while(q--)
        {
            int l,r,k;
            scanf("%d%d%d",&l,&r,&k);
            printf("%d\n",t[query(T[l],T[r+1],k)]);
        }
	}
}


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