BZOJ 3224 Tyvj 1728 普通平衡树 (Splay)

题目链接：
BZOJ 3224

题意：
让你实现一棵树，实现 插入， 删除，查询x数的排名，查询排名为x的数 ，求x的前驱(前驱定义为小于x，且最大的数)， 求x的后继(后继定义为大于x，且最小的数)的功能。

题解：
套平衡树Splay。

AC代码：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int Maxn = 100000 + 5, INF = 1 << 30;int fa[Maxn], ch[Maxn][2], key[Maxn], s[Maxn], root, tot, cnt[Maxn];void init(){    memset(fa, 0, sizeof fa);     memset(ch, 0, sizeof ch);    memset(key, 0, sizeof key);     memset(s, 0, sizeof s);    tot = root = 0;}void up(const int &u){        s[u] = s[ch[u][0]] + s[ch[u][1]] + cnt[u];}//x become the father of yvoid Rotate(const int &x, const int& d){    int y = fa[x];    ch[y][d^1] = ch[x][d];    if (ch[x][d]) fa[ch[x][d]] = y;    fa[x] = fa[y];    if (fa[x])    {        if (y == ch[fa[y]][0]) ch[fa[y]][0] = x;        else ch[fa[y]][1] = x;    }    ch[x][d] = y; fa[y] = x;    up(y); up(x);}//tag become x's fathervoid splay(const int &x, const int &tag){    while (fa[x] != tag)    {        int y = fa[x];        if (x == ch[y][0])        {            if (fa[y] != tag && y == ch[fa[y]][0]) Rotate(y,1);            Rotate(x,1);        }         else        {            if(fa[y] != tag && y == ch[fa[y]][1]) Rotate(y,0);                Rotate(x,0);            }        }        if (!tag) root = x;    }void ins(int &x, const int &val, const int &p){    if (!x) {        x = ++tot; key[x] = val; ch[x][0] = ch[x][1] = 0;        fa[x] = p; cnt[x] = s[x] = 1;     }     else{        int t = x;        if (val < key[t]) ins(ch[t][0],val,t);        else if (val > key[t]) ins(ch[t][1],val,t);        else ++cnt[x];        up(t);    }}void insert(const int &val){        ins(root, val, 0); splay(tot,0);}int find(int x, const int &val){    if (!x) return 0;    if (val < key[x]) return find(ch[x][0],val);    if (val > key[x]) return find(ch[x][1],val);    splay(x, 0);     return x;}    //delete rootvoid del(){    if (cnt[root] > 1)    {            --cnt[root]; --s[root]; return;    }    if( !ch[root][0] )    {        fa[ ch[root][1] ] = 0 ;        root = ch[root][1];    }    else    {        int cur = ch[root][0];        while( ch[cur][1] ) cur = ch[cur][1];        splay( cur , root );        ch[cur][1] = ch[root][1];        root = cur , fa[cur] = 0;        if ( ch[root][1] ) fa[ch[root][1]] = root;        up( root );    }}void Delete(const int& val){        int k = find(root,val);        if (k) del();}int Kth(int u, int k){        if (!u) return 0;        if (k <= s[ch[u][0]]) return Kth(ch[u][0],k);        if (k > s[ch[u][0]] + cnt[u]) return Kth(ch[u][1],k-s[ch[u][0]]-cnt[u]);        return key[u];}int Rank(int u, int val){    if (!u) return 0;    if (key[u] == val) return s[ch[u][0]] + 1;    if (key[u] < val) return s[ch[u][0]] + cnt[u] + Rank(ch[u][1],val);    else return Rank(ch[u][0],val);}int pred(int u, int val){    if (!u) return INF;    if (val <= key[u]) return pred(ch[u][0],val);    int ans = pred(ch[u][1],val);    if (ans == INF) ans = key[u];    return ans;}int succ(int u, int val){    if (!u) return INF;    if (val >= key[u]) return succ(ch[u][1],val);    int ans = succ(ch[u][0],val);    if (ans == INF) ans = key[u];    return ans;}int main(){    int t, op, x;    scanf("%d", &t);    init();    while (t--)    {        scanf("%d%d", &op, &x);        if (op == 1) insert(x); //插入         else if (op == 2) Delete(x); //删除         else if (op == 3) printf("%d\n", Rank(root,x)); //数x的排名         else if (op == 4) printf("%d\n", Kth(root,x)); //排名K的数         else if (op == 5)printf("%d\n", pred(root,x)); //前驱         else printf("%d\n", succ(root,x));//后继     }    return 0;}