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Feed the dogs
| Time Limit: 6000MS | Memory Limit: 65536K | |
| Total Submissions: 16655 | Accepted: 5203 |
Description
Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on
one line, numbered from 1 to n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should
be noted that Jiajia do not want to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may
intersect with each other.
Your task is to help Jiajia calculate which dog ate the food after each feeding.
Input
The first line contains n and m, indicates the number of dogs and the number of feedings.
The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.
Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.
You can assume that n<100001 and m<50001.
Output
Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.
Sample Input
7 2 1 5 2 6 3 7 4 1 5 3 2 7 1
Sample Output
3 2
题意：
给n个数和m个区间查询（每个查询区间之间可能有交集，但不会覆盖），每个查询输出序列的第k小的数。
思路：
据说可用各种树过~
伸展树，首先保存下所有查询，对查询的左边界按升序排序，然后对排序后的每个查询区间建立一颗独立的树，至于第k小，就可以直接计算出来；在为下一个查询区间建树时，如果跟前一次查询区间没有交集，就把整棵树删掉重新建；如果有交集，就可以把跟前一次查询区间没有交集的结点删掉，有交集的留下，再把当前查询区间没有交集的加上。
#include <cstdio>
#include <cstdlib>
#include <string>
#include <algorithm>
#include <string.h>
#include <cmath>
#include <iostream>
#define MAX(x, y) ((x)>(y)?(x):(y))
const int MAXN = 100100;
using namespace std;
typedef int Type;
Type a[MAXN], b[MAXN>>1];
struct Q
{
int x, y, z, id;
Q()
{
id = x = y = z = 0;
}
};
Q q[MAXN>>1];
bool cmp(Q q1, Q q2)
{
if(q1.x != q2.x) return q1.x < q2.x;
else return q1.y < q2.y;
}
typedef struct TREE
{
Type val;
TREE *fa, *l, *r;
int sz; //以该结点为根的树的总结点数
}Tree;
Tree *mark[MAXN]; //保存结点位置
struct SplayTree
{
public:
SplayTree()
{
rt = NULL;
inf = 1000000000;
}
void Push_up(Tree *T)
{
T->sz = (T->l ? T->l->sz : 0) + (T->r ? T->r->sz : 0) + 1;
}
void NewNode(Tree *pre, Tree *&T, Type v)
{
T = (Tree *)malloc(sizeof(Tree));
T->val = v;
T->sz = 1;
T->fa = pre;
T->l = T->r = NULL;
}
void Init()
{
NewNode(NULL, rt, -inf);
NewNode(rt, rt->r, inf);
rt->sz = 2;
}
void R_rotate(Tree *x)
{
Tree *y = x->fa;
Tree *z = y->fa;
Tree *k = x->r;
y->l = k;
x->r = y;
if(z){
if(y == z->l) z->l = x;
else z->r = x;
}
if(k) k->fa = y;
y->fa = x;
x->fa = z;
Push_up(y);
}
void L_rotate(Tree *x)
{
Tree *y = x->fa;
Tree *z = y->fa;
Tree *k = x->l;
y->r = k;
x->l = y;
if(z){
if(y == z->r) z->r = x;
else z->l = x;
}
if(k) k->fa = y;
y->fa = x;
x->fa = z;
Push_up(y);
}
//寻找第x个数的结点
Tree *FindTag(int x)
{
if(NULL == rt) return NULL;
Tree *p;
p = rt;
Type sum = (p->l ? p->l->sz : 0) + 1;
while(sum != x)
{
if(sum < x){
p = p->r;
x -= sum;
}
else p = p->l;
if(NULL == p) break;
sum = (p->l ? p->l->sz : 0) + 1;
}
return p;
}
void Splay(Tree *X, Tree *&T)
{
Tree *p, *end;
end = T->fa;
while(X->fa != end)
{
p = X->fa;
if(end == p->fa){ //p是根结点
if(X == p->l) R_rotate(X);
else L_rotate(X);
break;
}
//p不是根结点
if(X == p->l){
if(p == p->fa->l){
R_rotate(p); //LL
R_rotate(X); //LL
}
else{
R_rotate(X); //RL
L_rotate(X);
}
}
else{
if(p == p->fa->r){ //RR
L_rotate(p);
L_rotate(X);
}
else{ //LR
L_rotate(X);
R_rotate(X);
}
}
}
T = X;
Push_up(T);
}
void Insert(Type *A, int x, int y)
{
int i;
//考验指针技巧的代码
Tree **link, *p;
for(i = x; i <= y; i++){
link = &rt;
while(*link)
{
p = *link;
if(A[i] < p->val) link = &p->l;
else link = &p->r;
}
NewNode(p, *link, A[i]);
mark[i] = *link;
Splay(*link, rt);
}
}
void Delete(Type *A, int x, int y)
{
int i;
Tree *t;
for(i = x; i <= y; i++){
t = mark[i];
Splay(t, rt);
t = rt->l;
while(t->r) t = t->r;
Splay(t, rt->l);
t = rt;
rt = rt->l;
rt->r = t->r;
free(t);
rt->r->fa = rt;
rt->fa = NULL;
Push_up(rt); //切记更新sz
}
}
void Query(int id, int x)
{
x++; //自增1，因为有个-oo结点
Tree *p;
p = rt;
int sum = (p->l ? p->l->sz : 0) + 1;
while(sum != x)
{
if(sum < x){
p = p->r;
x -= sum;
}
else p = p->l;
if(NULL == p) break;
sum = (p->l ? p->l->sz : 0) + 1;
}
b[id] = p->val;
}
void Show()
{
InOrder(rt);
printf("\n");
}
void InOrder(Tree *T)
{
if(NULL == T) return;
InOrder(T->l);
printf("%d ", T->val);
InOrder(T->r);
}
void Free()
{
FreeTree(rt);
}
void FreeTree(Tree *T)
{
if(NULL == T) return;
FreeTree(T->l);
FreeTree(T->r);
free(T);
}
private:
Type inf;
Tree *rt;
};
SplayTree spl;
int main()
{
//freopen("in.txt","r",stdin);
int n, m, i;
while(scanf("%d%d", &n, &m) == 2)
{
for(i = 1; i <= n; i++)
scanf("%d", a + i);
for(i = 1; i <= m; i++){
scanf("%d%d%d", &q[i].x, &q[i].y, &q[i].z);
q[i].id = i; //把位置记下方便输出
}
sort(q + 1, q + m + 1, cmp);
for(i = 1; i <= m; i++){
if(q[i].x <= q[i-1].y){ //有交集
spl.Delete(a, q[i-1].x, q[i].x - 1);
spl.Insert(a, q[i-1].y + 1, q[i].y);
}
else{
spl.Free(); //把整棵树删掉
spl.Init();
spl.Insert(a, q[i].x, q[i].y);
}
spl.Query(q[i].id, q[i].z);
}
spl.Free();
for(i = 1; i <= m; i++)
printf("%d\n", b[i]);
}
return 0;
}
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