HDU 3564 Another LIS splay（水

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题意：

给定一个空序列

插入n个数（依次插入 1、2、3、4··n）

下面n个数表示i插在哪个位置。

每插入一个数后输出这个序列的lis

然后。。。

因为每次插入的数都是当前序列最大的数

所以不会影响后面的数的dp值

那么这个位置的dp值就是插入位置的前面最大dp值+1

然后输出这个序列最大的dp值。

思路：

splay。。。

Q：为何这题需要用splay，不是简单的线段树吗

A: 因为我智商不够想不出线段树怎么写。。

#include <cstdio>#include <iostream>#include <cstring>#include <queue>#include <algorithm>#include <map>#include <cmath>template <class T>inline bool rd(T &ret) {char c; int sgn;if(c=getchar(),c==EOF) return 0;while(c!='-'&&(c<'0'||c>'9')) c=getchar();sgn=(c=='-')?-1:1;ret=(c=='-')?0:(c-'0');while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');ret*=sgn;return 1;}template <class T>inline void pt(T x) {    if (x <0) {        putchar('-');        x = -x;    }    if(x>9) pt(x/10);    putchar(x%10+'0');}using namespace std;inline int Mid(int a,int b){return (a+b)>>1;}#define N 100010#define L(x) tree[x].ch[0]#define R(x) tree[x].ch[1]#define Siz(x) tree[x].siz#define Father(x) tree[x].fa#define Max(x) tree[x].max#define Val(x) tree[x].val#define Pt(x) tree[x].pt()struct node{    int ch[2], siz, fa;    int max, val;    void pt(){printf("val:%d max:%d siz:%d fa:%d{%d,%d}\n", val,max,siz,fa,ch[0],ch[1]);}}tree[N*2];int tot, root;void Newnode(int &id, int val, int fa, int siz = 1){    id = ++tot;    L(id) = R(id) = 0;    Father(id) = fa;    Siz(id) = siz;    Max(id) = Val(id) = val;}void push_up(int id){    Siz(id) = Siz(L(id)) + Siz(R(id)) +1;    Max(id) = max(Max(R(id)), Max(L(id)));    Max(id) = max(Val(id), Max(id));}void push_down(int id){}void Rotate(int id, int kind){    int y = Father(id);    push_down(y); push_down(id); //here    tree[y].ch[kind^1] = tree[id].ch[kind];    Father(tree[id].ch[kind]) = y;    if(Father(y))        tree[Father(y)].ch[R(Father(y))==y] = id;    Father(id) = Father(y);    Father(y) = id;    tree[id].ch[kind] = y;    push_up(y);}void splay(int id, int goal){    push_down(id);    while(Father(id) != goal){        int y = Father(id);        if(Father(y) == goal)            Rotate(id, L(y)==id);        else        {            int kind = L(Father(y)) == y;            if(tree[y].ch[kind] == id)            {                Rotate(id, kind^1);                Rotate(id, kind);            }            else            {                Rotate(y, kind);                Rotate(id,kind);            }        }    }    push_up(id);    if(goal == 0)root = id;}int Get_kth(int kth, int sor){//找到在sor后面的第k个数    push_down(sor);    int id = sor;    while(Siz(L(id)) != kth){        if(Siz(L(id)) > kth)            id = L(id);        else        {            kth -= (Siz(L(id))+1);            id = R(id);        }        push_down(id);    }    return id;}void init(){Father(0) = L(0) = R(0) = Siz(0) = 0;Max(0) = 0;tot = 0;Newnode(root, 0, 0);Newnode(R(root), 0, root);push_up(root);}void debug(int x){printf("%d:\n", x);Pt(x);if(L(x))debug(L(x));if(R(x))debug(R(x));}void insert(int pos){splay(1, 0);int u = Get_kth(pos, 1);//if(pos == 2){cout<<"=="; debug(root);}int v = Get_kth(pos+1, 1);splay(u, 0);splay(v, root);//if(pos == 2){cout<<"=="; debug(root);}Newnode(L(v), max(Val(root), Max(L(root))) +1, v);push_up(v);push_up(u);//printf("[%d,%d]\n", u, v);}int n;int main() {int T, Cas = 1; cin>>T;    while(T--){    rd(n);    init();    //debug(root);    printf("Case #%d:\n", Cas++);    for(int i = 1, m; i <= n; i++){    rd(m);    insert(m);    //printf("id:%d, pos:%d\n", i, m);    debug(root);    pt(Max(root)); putchar('\n');    splay(tot, 0);    //puts("================");debug(root);    }/**/        puts("");    }    return 0;}/*170 1 1 1 0 4 1*/

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