数据结构三
左偏树可能各个方面都是最优的,可能他的最大缺点是插入操作O(n),但是综合并操作与代码难度考虑还是不错的选择
至于他的已知节点删除和变种没实现,待以后用的话再补上吧,觉得几乎没用处?(不敢说XD...
/**************************************** * LeftHeap.h * Describe: 可并优先队列——左偏树 * Created: 2010-5-27 * CopyRight 2010,Leek * ******************************************/ #include <iostream> #include <vector> #include <queue> using namespace std; typedef int Type; typedef struct LNode { Type key; int dis; LNode *left; LNode *right; LNode (Type key) { dis = 0; left = NULL; right = NULL; this->key = key; } }LNode; typedef LNode *PLNode; void Swap (int &a, int &b) { a ^= b, b ^= a, a ^= b; } void Swap (PLNode &a, PLNode &b) { PLNode tmp = a; a = b, b = tmp; } PLNode LHeap_Merge (PLNode a,PLNode b) { if (a == NULL) return b; if (b == NULL) return a; if (a->key > b->key) Swap (a->key,b->key); a->right = LHeap_Merge (a->right, b); if (a->left == NULL || a->right->dis > a->left->dis) Swap (a->right,a->left); if (a->right == NULL) a->dis = 0; else a->dis = a->right->dis + 1; return a; } PLNode LHeap_Build (queue<PLNode> &a) { PLNode tmp = a.front(); while (a.size() > 1) { a.pop(); tmp = LHeap_Merge (tmp,a.front()); a.push(tmp); a.pop(); tmp = a.front(); } return a.front(); } PLNode LHeap_Insert (Type key,PLNode root) { PLNode tmp = new LNode(key); return LHeap_Merge (tmp,root); } Type LHeap_Delete_Min (PLNode &root) { PLNode tmp = root; root = LHeap_Merge (root->left, root->right); Type top = tmp->key; delete tmp; return top; }
觉得建树和并操作值得思考多少有点启发性吧
顺便贴上败者树的模板 今天复习结束看漫画去
/**************************************** * Loser.h * Describe: 败者树 * Created: 2010-5-27 * CopyRight 2010,Leek * ******************************************/ #include <iostream> using namespace std; const int N = 1 << 16; //N大于n,并为2的次方 const int n = 234; typedef int Type; Type f[N * 2]; Type a[n]; void BuildTree (int n, int f[2 * N]) //建树a存在f N-N+n { memcpy (f + N,a,sizeof (a[0]) *n); for (int i = N,j = N + n - 1; i > 1;i >>= 1,j >>= 1) { for (int k = i; k < j;k += 2) f[k >> 1] = f[k] > f[k + 1] ? f[k] : f[k + 1]; //求得是最大值,可改为最小值 if (!(j&1)) f[j >> 1] = f[j]; } } int findmax (int s, int t) //查询s t中的最大 最小值 { int r = f[s] >= f[t]? s: t; for (int i = s,j = t; i < j; i >>= 1, j >>= 1) { if (!(i & 1) && i + 1 < j && f[i + 1] > f[r]) r = i + 1; if ((j & 1) && j - 1 > i && f[j - 1] > f[r]) r = j - 1; } return f[r]; //以下程序可求a【s t】最大最小值的下标 while (r < N) if (f[r] == f[r << 1]) r <<= 1; else if (f[r] == f[(r << 1) + 1]) r = (r << 1) + 1; }
数据结构二
关于平衡二叉树只会AVL和傻X树:( 由于AVL各方面都不如SBT 所以只整理SBT算了。
但是杯具的是把以前整理写的代码等Ctrl+A Shift+Delete了...
这个是模仿别人的重新写一个,其中删除是最有技巧结合平衡操作很精妙
/**************************************** * SBT.h * Describe: SBT平衡二叉树 * Created: 2010-5-26 * CopyRight 2010,Leek * ******************************************/ #include <iostream> using namespace std; typedef int Type; // SBT存储结构 class SBTNode { public: SBTNode *child[2]; SBTNode *parent; Type key; int size; SBTNode (Type key,int size); }; typedef SBTNode *SBTree; enum {Left,Right=1}; SBTNode::SBTNode (Type key,int size) { this->key = key; this->size = size; child[Left] = child[Right] = parent = NULL; } void SBT_Rotate (SBTree &root, bool num) { //num == Left 右旋num == Right左旋 SBTNode *tmp = root->child[!num]; tmp->parent = root->parent; if (tmp->child[num] != NULL) tmp->child[num]->parent = root; root->child[!num] = tmp->child[num]; tmp->child[num] = root; tmp->size = root->size; root->size = root->child[Left]->size + root->child[Right]->size + Right; root = tmp; } void SBT_Maintain (SBTree &root, bool num) { //人字型数据退化,但是其他数据速度都快 if (root->child[num]->child[num]->size > root->child[!num]->size) SBT_Rotate (root, !num); } void SBT_Insert (SBTree &root, SBTNode *x) { if (root == NULL) root = x; else { root->size++; x->parent = root; SBT_Insert (root->child[x->key > root->key], x); SBT_Maintain (root, x->key > root->key); } } SBTNode *SBT_Search (SBTree &root,Type key) { return root == NULL || root->key == key? root: SBT_Search (root->child[key > root->key], key); } SBTNode *SBT_Delete (SBTree &root, Type key) { if (root == NULL) return NULL; root->size--; if (root->key == key || root->child[key > root->key] == NULL) { SBTNode *tmp; if (root->child[Left] == NULL || root->child[Right] ==NULL) { tmp = root; root = root->child[root->child[Left] == NULL]; if (root != NULL) root->parent = tmp->parent; } else { tmp = SBT_Delete (root->child[Right], key - Right); root->key = tmp->key; } return tmp; } else return SBT_Delete (root->child[key > root->key], key); } SBTNode *SBT_Pred(SBTree root, Type key) { if (root == NULL) return NULL; if(key <= root->key) return SBT_Pred(root->child[Left],key); else { SBTNode *tmp=SBT_Pred(root->child[Right],key); return tmp != NULL?tmp:root; } } SBTNode *SBT_Succ (SBTree &root,Type key) { if (root == NULL) return NULL; if (key >= root->key) return SBT_Succ(root->child[Right],key); SBTNode *tmp = SBT_Succ(root->child[Left], key); return tmp != NULL? tmp:root; } SBTNode *SBT_Select(SBTree root, int i) { int r = root->child[Left]->size+Right; if (i == r) return root; else return SBT_Select(root->child[i>r],i>r?i-r:i); } int SBT_Rank(SBTree root, Type key) { if (root == NULL) return Left; if (root->key == key) return root->child[Left]->size+Right; else if(key < root->key) return SBT_Rank(root->child[Left],key); else { int r = SBT_Rank(root->child[Right],key); return r == Left? Left: r + root->child[Left]->size + Right; } }
等有时间改成模板实现吧
第一次应聘实习 | 排列与阶乘进制
昨天去ND面试 (车上收到有道短信 说下午视频答疑,估计也没什么问题问的就算了 .主考官直接问我什么时候有时间过去 (囧。然后说暑假吧,直接被他否决了 ,说是不能那么晚,现在就得去,平时没时间可以请假一个星期有到三四天就行了...就这样默默地答应了。之后回来打电话给我说明天要过去签协议,补贴方面800保障,出勤好的话每个月加80(可是网上说是1100,这里怎么变了呢 ,他解释什么税收管理费等(= =不是2k以上才要税收吗,显然地有猫腻...被人坑的感觉自然不爽加上路途漫漫(温泉路口 所以今天一大早打电话过去说不去了 酱紫第一次很山寨的笔试面试经历结束了...
回来继续复习和整理算法。发现模板的《排列与阶乘进制》有猫腻,大牛的代码写的很玄乎看不出是不是真的错了(代码不能多几个注释嘛,然后按自己的想法改动了跑了一下果然假想没错
现在就说说Coding吧,之前排列与顺序等问题直接都是土土地写地,看了阶乘进制赞叹Knuth大师的伟大(... 然后再看逆过程无比纠结(反正死活看不懂喽,然后偷偷暗想他Coding有问题吧,越看越有问题(= = 鄙视一下自己,经过简单数据验证还好真的是模板的问题(XD
原先的逆过程:
void back(__int64 perm) { int i, j, k, p[N + 1] = {0},mark[N + 1] = {0}; i = 1; while (perm) { p[i] = perm % (i + 1); perm /= ++i; } for (i = n-1; i >= 1; --i) { j = n,k = 0; while (k <= p[i]) { if (!mark[j++]) k++; } j++; a[j] = i + 1; mark[j] = 1; } j = 1; while (mark[j]) j++; a[j] = 1; } 修改后代码: void back(__int64 perm) { int i, j, k, p[N + 1] = {0},mark[N + 1] = {0}; i = 1; while (perm) { p[i] = perm % (i + 1); perm /= ++i; } for (i = n-1; i >= 1; --i) { j = 1,k = 0; while (k <= p[i]) { if (!mark[j++]) k++; } j--; a[n-i] = j; mark[j] = 1; } j = 1; while (mark[j]) j++; a[N] = j; }
顺便贴上转化为阶乘进制代码: int KnuthPEncode() { int i,j; int place = n,e = 0; for (i = 1; i <= n; ++i) { int t = a[i]; e *= place--; for (j = i+1; j <= n ; ++j) a[j] < t && e++; } return e; }
数据结构一
由于应聘网龙实习生耽搁了,先把整理好的基础部分贴出来
据说ND招实习生就是招廉价劳动力?随便吧,反正不打算能做成怎么样可以向大牛们学点东西才是关键。在想要是实习住宿问题怎么解决?YOYO大人赞助吧?嗯..
不扯淡了直接晒代码
//*************二叉堆相关操作********************* void Sink(int p) { int q = p << 1, tmp = heap[p]; while (q <= hs) { if (q < hs && heap[q + 1] < heap[q]) ++q; if (heap[p] < heap[q]) break; heap[q] = heap[p]; p = q; q <<= 1; } heap[p] = tmp; } int Delete_MinI() { int r = heap[1]; heap[1] = heap[hs--]; Sink(1); return r; } int Delete_MinII() { heap[1] ^= heap[hs], heap[hs] ^= heap[1], heap[1] ^= heap[hs--]; Sink(1); return heap[hs + 1]; } void Swim(int p) { int q = p >> 1, tmp = heap[p]; while(q && tmp < heap[q]) { heap[p] = heap[q]; p = q; q >>= 1; } heap[p] = tmp; } void Insert(int p) { heap[++hs] = p; Swim(hs); } void Build() { for (int i = hs / 2; i ; --i) Sink(i); } /**********并查集************************/ void Make_Set(int x) { rank[x] = 0; p[x] = x; } int Find_Set(int x) { int px = x, tmp; while (px != p[px]) px = p[px]; while (x != px) { tmp = p[x]; p[x] = px; x = tmp; } return px; } void Union_Set(int a,int b) { a = Find_Set(a); b = Find_Set(b); if (rank[a] < rank[b]) { p[a] = b; } else { p[b] = a; if (rank[a] == rank[b]) ++rank[a]; } } /************哈希表********************/ //质数1811 5087 10657 50503 100109 500119 int hash[Size * Size][2]; int first[Size][Size]; int next[Size * Size]; int entry = 1; /*********************/ // 值太大二维存储 ps:从没用过 似乎没必要? /*********************/ int Insert_Find_Hash() { long long a, b; a = 前半部分的值(可按位存储);b = 后半部分; c = a % Size; d = b % Size; e = first[c][d]; while (e != 0) { if (hash[e][0] == a && hash[e][1] == b) return 0; e = next[e]; } hash[entry][0] = a; hash[entry][1] = b; next[entry] = first[c][d]; first[c][d] = entry; entry ++; return 1; } bool Insert_Find_Hash(Type x) { unsigned int key; key = Hash(x); p = first[key]; while (p) { if (hash[p] == x) return false; p = next[p]; } hash[entry] = key; next[entry] = first[key]; first[key] = entry; entry ++; return true; } //字符串用的Hash函数 unsigned int SDBMHash(char *str) { unsigned int hash = 0; unsigned int len = 0; while (*str) { // hash = 65599*hash + (*str++); hash = (*str++) + (hash << 6) + (hash << 16) - hash; ++len; } hash = (len) + (hash << 6) + (hash << 16) - hash; return (hash & 0x7FFFFFFF); } /**********Good to the last code************/
下期:平衡二叉树