最近“hiho一下”出了平衡树专题，这周的Splay一直出现RE，应该删除操作指针没处理好，还没找出原因。
不过其他操作运行正常，尝试用它写了一道之前用set做的平衡树的题http://codeforces.com/problemset/problem/675/D，运行效果居然还挺好的，时间快了大概10%，内存少了大概30%。
1 #include <cstdio> 2 #include <cstring> 3 #include <string> 4 #include <cstdlib> 5 #include <cctype> 6 #include <cmath> 7 #include <algorithm> 8 #include <vector> 9 #include <map> 10 #include <set> 11 #include <stack> 12 #include <queue> 13 #include <assert.h> 14 #define FREAD(fn) freopen((fn), "r", stdin) 15 #define RINT(vn) scanf("%d", &(vn)) 16 #define PINT(vb) printf("%d", vb) 17 #define RSTR(vn) scanf("%s", (vn)) 18 #define PSTR(vn) printf("%s", (vn)) 19 #define CLEAR(A, X) memset(A, X, sizeof(A)) 20 #define REP(N) for(i=0; i<(N); i++) 21 #define REPE(N) for(i=1; i<=(N); i++) 22 #define pb(X) push_back(X) 23 #define pn() printf("\n") 24 using namespace std; 25 const int MAX_N = 100005; 26 const int MAX_K = 0x7fffffff; 27 const int MIN_K = 0; 28 29 int a[MAX_N]; 30 int n; 31 int i; 32 map<int, bool> left, right;//iostream里有和left, right冲突的命名! 33 34 struct Node 35 { 36 int k; 37 Node *l, *r, *p; 38 Node():k(-1), l(NULL), r(NULL), p(NULL){} 39 Node(int kk, Node* pp):k(kk), l(NULL), r(NULL), p(pp){} 40 ~Node(){ 41 l = r = p = NULL; 42 } 43 }; 44 45 struct Splay 46 { 47 Node* root; 48 Node* _hot; 49 Splay():root(NULL), _hot(NULL){} 50 Splay(int k):root(new Node(k, NULL)), _hot(root){} 51 52 void release(Node* cur){//释放子树cur的空间 53 if(cur == NULL) return ;//空树 54 release(cur->l); 55 cur->l = NULL; 56 release(cur->r); 57 cur->r = NULL; //不用加吧，cur马上就销毁了啊 58 //printf("deleted %d\n", cur->k); 59 60 delete cur; 61 return ; 62 } 63 ~Splay(){ 64 release(root); 65 root = NULL; 66 } 67 void zig(Node* cur){ 68 if(cur == NULL) return ; 69 Node* v = cur->l; 70 if(v == NULL) return ; 71 Node* g = cur->p; 72 73 v->p = g; 74 if(g != NULL) 75 //祖先g与v连接 76 (cur == g->l) ? g->l = v : g->r = v; 77 78 //v与cur孩子过继 79 cur->l = v->r; 80 if(cur->l != NULL) cur->l->p = cur; 81 82 //v与cur角色转换 83 cur->p = v; 84 v->r = cur; 85 if(cur == root) root = v; 86 //printf("%d zigged\n", cur->k); 87 } 88 void zag(Node* cur){ 89 if(cur == NULL) return ; 90 Node* v = cur->r; 91 if(v == NULL) return ; 92 Node* g = cur->p; 93 //printf("g=%d cur=%d v=%d\n", g->k, cur->k, v->k); 94 95 v->p = g; 96 if(g != NULL) 97 (cur == g->l) ? g->l = v : g->r = v; 98 99 cur->r = v->l; 100 if(cur->r != NULL) cur->r->p = cur; 101 102 cur->p = v; 103 v->l = cur; 104 if(cur == root) root = v; 105 //printf("%d zagged\n", cur->k); 106 } 107 void splay(Node* x, Node* f){// make x become f's child 108 if(x == NULL) return ; 109 while(x->p != f){//逐步双层伸展 110 Node* p = x->p; 111 if(p == NULL) return ; 112 if(p->p == f) 113 (x == p->l) ? zig(p) : zag(p); 114 else{ 115 Node* g = p->p; 116 if(g == NULL) return ; 117 if(g->l == p){ 118 if(p->l == x){ 119 zig(g); zig(p); 120 }else{ 121 zag(p); zig(g); 122 } 123 }else{ 124 if(p->l == x){ 125 zig(p); zag(g); 126 }else{ 127 zag(g); zag(p); 128 } 129 } 130 } 131 } 132 } 133 Node* search(Node* cur, int k){//在cur子树中查找关键码k 134 if(cur == NULL) return _hot;//查找失败还伸展吗？暂不伸展，待决定插入后再将新插入的节点伸展 135 if(cur->k == k){//查找成功 136 //printf("has %d\n", cur->k); 137 splay(cur, NULL);//将目标节点伸展至根 138 return cur; 139 } 140 _hot = cur;//需要深入子树查找 141 return (k < cur->k) ? search(cur->l, k) : search(cur->r, k); 142 } 143 Node* insert(Node* cur, int k){//将关键码k插入cur子树 144 if(cur == NULL){//找到目标插入位置 145 cur = new Node(k, _hot); 146 //printf("%d %d\n", _hot->k, k); 147 (k < _hot->k) ? _hot->l=cur : _hot->r=cur; 148 _hot = cur; 149 //printf("create %d\n", cur->k); 150 splay(cur, NULL);//将目标节点伸展至树根 151 return cur; 152 } 153 assert(cur); 154 _hot = cur;//进入子树 155 //printf("enter %d\n", cur->k); 156 return (k < cur->k) ? insert(cur->l, k) : insert(cur->r, k);//assert:关键码互异 157 } 158 Node* prev(int k){//寻找关键码k的中序前驱 159 splay(search(root, k), NULL);//将k伸展至树根 160 Node* cur = root->l;//根节点的左子树 161 //assert(cur); 162 if(!cur) return NULL; 163 while(cur->r != NULL) cur = cur->r;//前驱必然为左子树的最右节点 164 return cur; 165 } 166 Node* succ(int k){//寻找关键码k的中序后继, assert:k一定存在 167 splay(search(root, k), NULL); 168 Node* cur = root->r; 169 //assert(cur); 170 if(!cur) return NULL; 171 while(cur->l != NULL) cur = cur->l; 172 return cur; 173 } 174 void deleteK(int k){//删除关键码k 175 Node* p = prev(k); 176 Node* s = succ(k); 177 splay(p, NULL); 178 splay(s, p); 179 Node* q = s->l; 180 s->l = NULL;//解除父子关系 181 release(q);//释放子树空间，这里只有一个节点k 182 } 183 void deleteInterval(int a, int b){//删除区间[a,b]内的关键码 184 Node* pa = search(root, a);//pa为最后一个被访问的节点，必不空 185 assert(pa); 186 if(pa->k != a) pa = insert(pa, a);//查找失败，插入 187 //printf("pa->k = a = %d\n", pa->k); 188 189 Node* pb = search(root, b); 190 assert(pb); 191 //printf("pb->k = b = %d\n", pb->k); 192 if(pb->k != b) pb = insert(pb, b); 193 194 Node* p = prev(a); 195 assert(p); 196 Node* s = succ(b);//assert: p, s not null 197 assert(s); 198 //printf("prev %d succ %d\n", p->k, s->k); 199 splay(p, NULL); 200 //printf("%d splayed\n", p->k); 201 splay(s, p); 202 //printf("%d splayed\n", s->k); 203 Node* q = s->l; 204 _hot = s; 205 release(q);//释放子树空间 206 s->l = NULL; 207 } 208 }; 209 210 int main() 211 { 212 FREAD("675d.txt"); 213 RINT(n); 214 REP(n) RINT(a[i]); 215 Splay mySplay(a[0]); 216 for(i=1; i<n; i++){ 217 // Node* p = mySplay.search(mySplay.root, a[i]);//必然失败 218 // if(p->k > a[i]) 219 // printf("%d\n", mySplay.prev(p->k)); 220 // else printf("%d\n", p->k); 221 int ans = 0; 222 Node* q = mySplay.insert(mySplay.root, a[i]); 223 Node* p = mySplay.prev(a[i]); 224 if(p && right.count(p->k) == 0){ 225 right[p->k] = 1;//前驱没有右孩子 226 ans = p->k; 227 }else{ 228 Node* s = mySplay.succ(a[i]); 229 if(s && left.count(s->k) == 0){ 230 left[s->k] = 1; 231 ans = s->k; 232 } 233 } 234 printf("%d\n", ans); 235 } 236 return 0; 237 }