poj 2418 Hardwood Species
原题链接:http://poj.org/problem?id=2418
简单题。。
平衡树,写了个模板。。动态分配内存确实很慢。。。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<algorithm> 2 #include<iostream> 3 #include<string> 4 #include<cstdlib> 5 #include<cstring> 6 #include<cstdio> 7 using std::string; 8 template<typename T> 9 class sb_tree{ 10 private: 11 struct Node{ 12 T data; 13 int s, c; 14 Node *ch[2]; 15 Node(const T&d) :s(1), c(1), data(d){} 16 Node() :s(0), c(0){ ch[0] = ch[1] = this; } 17 inline void push_up(){ 18 s = ch[0]->s + ch[1]->s + c; 19 } 20 inline int cmp(const T&v) const{ 21 return v == data ? -1 : v > data; 22 } 23 }*null, *root; 24 inline void rotate(Node* &x, int d){ 25 Node *k = x->ch[!d]; 26 x->ch[!d] = k->ch[d]; 27 k->ch[d] = x; 28 k->s = x->s; 29 x->push_up(); 30 x = k; 31 } 32 inline void Maintain(Node* &x, int d){ 33 if (x->ch[d] == null) return; 34 if (x->ch[d]->ch[d]->s > x->ch[!d]->s){ 35 rotate(x, !d); 36 } else if (x->ch[d]->ch[!d]->s > x->ch[!d]->s){ 37 rotate(x->ch[d], d), rotate(x, !d); 38 } else { 39 return; 40 } 41 Maintain(x, 0), Maintain(x, 1); 42 } 43 inline void insert(Node *&x, const T&data){ 44 if (!x->s){ 45 x = new Node(data); 46 x->ch[0] = x->ch[1] = null; 47 } else{ 48 x->s++; 49 int d = x->cmp(data); 50 if (-1 == d){ 51 x->c++; 52 return; 53 } 54 insert(x->ch[d], data); 55 x->push_up(); 56 Maintain(x, d); 57 } 58 } 59 inline void del(Node *&x, const T&data){ 60 if (!x->s) return ; 61 x->s--; 62 int d = x->cmp(data); 63 if (-1 == d){ 64 if (x->c > 1){ 65 x->c--; 66 return; 67 } else if (!x->ch[0]->s || !x->ch[1]->s){ 68 x = x->ch[0]->s ? x->ch[0] : x->ch[1]; 69 } else { 70 Node *ret = x->ch[1]; 71 for (; ret->ch[0]->s; ret = ret->ch[0]); 72 del(x->ch[1], x->data = ret->data); 73 } 74 } else { 75 del(x->ch[d], data); 76 } 77 if (x->s) x->push_up(); 78 } 79 void clear(Node* &x){ 80 if (x->s) clear(x->ch[0]), clear(x->ch[1]), delete x; 81 } 82 public: 83 sb_tree() :null(new Node), root(null){} 84 ~sb_tree(){ clear(root), delete null; } 85 inline void clear(){ clear(root), root = null; } 86 inline void insert(const T&data){ 87 insert(root, data); 88 } 89 inline void del(const T&data){ 90 del(root, data); 91 } 92 inline bool find(const T&data){ 93 Node *x = root; 94 while (x->s && x->data != data) x = x->ch[x->data < data]; 95 return x->s; 96 } 97 inline const T&kth(int k){ 98 Node *x = root; 99 for (int t = 0; x != null;){ 100 t = x->ch[0]->s; 101 if (k <= t) x = x->ch[0]; 102 else if (t + 1 <= k && k <= t + x->c) break; 103 else k -= t + x->c, x = x->ch[1]; 104 } 105 return x->data; 106 } 107 inline int rank(const T&data){ 108 Node *x = root; 109 int t = 0, cur = 0; 110 for (cur = 0; x != null;){ 111 t = x->ch[0]->s; 112 if (data == x->data) break; 113 else if (data < x->data) x = x->ch[0]; 114 else cur += t + x->c, x = x->ch[1]; 115 } 116 return cur + t + 1; 117 } 118 inline const T&operator[](int k){ 119 return kth(k); 120 } 121 inline const T&preorder(const T&data){ 122 Node *x = root, *y = 0; 123 while (x->s){ 124 if (x->data < data) y = x, x = x->ch[1]; 125 else x = x->ch[0]; 126 } 127 return y ? y->data : data; 128 } 129 inline const T&successor(const T&data){ 130 Node *x = root, *y = 0; 131 while (x->s){ 132 if (data < x->data) y = x, x = x->ch[0]; 133 else x = x->ch[1]; 134 } 135 return y ? y->data : data; 136 } 137 inline int size(){ return root->s; } 138 inline void query(Node *x, int cnt){ 139 if (x != null){ 140 query(x->ch[0], cnt); 141 string str = x->data; 142 printf("%s %.4lf\n", str.c_str(), (double)x->c * 100 / cnt); 143 query(x->ch[1], cnt); 144 } 145 } 146 inline void query(){ 147 int cnt = size(); 148 query(root, cnt); 149 } 150 }; 151 int main(){ 152 #ifdef LOCAL 153 freopen("in.txt", "r", stdin); 154 freopen("out.txt", "w+", stdout); 155 #endif 156 char buf[100]; 157 sb_tree<string> ans; 158 while (gets(buf)) ans.insert(buf); 159 ans.query(); 160 return 0; 161 }
如果不想写平衡树,就写个快排吧。。注意用c提交:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 typedef char State[33]; 5 State st[1000010]; 6 int cmp(const char* src, const char *_src){ 7 return strcmp((char *)src, (char *)_src); 8 } 9 int main(){ 10 State ret, buf; 11 int i, k, cnt = 0; 12 while (gets(ret)) strcpy(st[cnt++], ret); 13 qsort(st, cnt, sizeof(st[0]), cmp); 14 i = 0; 15 while (i < cnt){ 16 strcpy(buf, st[i]), k = 0; 17 for (; 0 == strcmp(buf, st[i]); i++) k++; 18 printf("%s %.4lf\n", buf, (double)k * 100 / cnt); 19 } 20 return 0; 21 }
By: GadyPu 博客地址:http://www.cnblogs.com/GadyPu/ 转载请说明