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刚学完后缀数组求回文串的瑶瑶(tsyao)想到了另一个问题:如果能够对字符串做一些修改,怎么在每次询问时知道以某个字符为中心的最长回文串长度呢?因为瑶瑶整天只知道LOL,当他知道自己省选成绩的时候就天天在LOL,导致现在的她实在是太弱了,根本解决不了这个问题,于是就来找你帮忙,么么哒~你就帮帮她吗
第一行为一个长度不超过100000字符串s作为初始字符串。第二行一个正整数n,表示操作/询问的个数。接下来n行,每行有如下几种可能出现的操作/询问:
Insert a x 在a处字符的后面插入一个字符x
Delete a 把a处字符删除
Update a x 把a处字符改为x
Query a 查询以a为中心的最长回文串长度
对于每个询问,输出得到的最长回文串长度
题目大意:略。
思路:
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搬一下官方(?)题解:http://tsyao.tk/archives/94
SPLAY保存改区间从左向右的hash值和从右向左的hash值,对于每个询问,二分字符串长度。总的时间复杂度是O(qlognlogn)
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处理技巧:头尾加上一个空字符可以不用考虑边界的情况。
代码(1932MS):
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 typedef long long LL; 7 8 const int MAXN = 150010; 9 const int MOD = 1e8 + 7; 10 const LL seed = 131; 11 12 LL base[MAXN]; 13 char s[MAXN], op[10]; 14 int n, m; 15 16 void initBase(int n = 150000) { 17 base[0] = 1; 18 for(int i = 1; i <= n; ++i) base[i] = base[i - 1] * seed % MOD; 19 } 20 21 struct SplayTree { 22 struct Node { 23 int size, lhash, rhash; 24 char c; 25 Node *fa, *ch[2]; 26 }; 27 Node statePool[MAXN], *nil, *root; 28 int stk[MAXN], top; 29 int ncnt; 30 31 SplayTree() { 32 nil = statePool; 33 } 34 35 void init() { 36 ncnt = 1; 37 top = 0; 38 } 39 40 Node* new_node(char v, Node* f) { 41 Node* t; 42 if(top) t = &statePool[stk[--top]]; 43 else t = &statePool[ncnt++]; 44 t->size = 1; 45 t->lhash = t->rhash = t->c = v; 46 t->ch[0] = t->ch[1] = nil; 47 t->fa = f; 48 return t; 49 } 50 51 void del_node(Node* &x) { 52 stk[top++] = x - statePool; 53 x = nil; 54 } 55 56 void update(Node* x) { 57 int s0 = x->ch[0]->size, s1 = x->ch[1]->size; 58 x->size = s0 + s1 + 1; 59 x->lhash = (x->ch[0]->lhash * base[s1 + 1] + x->c * base[s1] + x->ch[1]->lhash) % MOD; 60 x->rhash = (x->ch[1]->rhash * base[s0 + 1] + x->c * base[s0] + x->ch[0]->rhash) % MOD; 61 } 62 63 void rotate(Node* x) { 64 Node* y = x->fa; 65 int t = (y->ch[1] == x); 66 y->fa->ch[y->fa->ch[1] == y] = x; x->fa = y->fa; 67 y->ch[t] = x->ch[t ^ 1]; x->ch[t ^ 1]->fa = y; 68 x->ch[t ^ 1] = y; y->fa = x; 69 update(y); 70 } 71 72 void splay(Node* x, Node* f) { 73 while(x->fa != f) { 74 if(x->fa->fa == f) rotate(x); 75 else { 76 Node *y = x->fa, *z = y->fa; 77 if((z->ch[1] == y) == (y->ch[1] == x)) rotate(y); 78 else rotate(x); 79 rotate(x); 80 } 81 } 82 update(x); 83 if(x->fa == nil) root = x; 84 } 85 86 Node* kth(int k) { 87 Node* x = root; 88 while(true) { 89 int t = x->ch[0]->size + 1; 90 if(t == k) break; 91 if(t > k) x = x->ch[0]; 92 else x = x->ch[1], k -= t; 93 } 94 return x; 95 } 96 97 void build(Node* &x, Node* f, int l, int r) { 98 int mid = (l + r) >> 1; 99 x = new_node(s[mid], f); 100 if(l < mid) build(x->ch[0], x, l, mid - 1); 101 if(mid < r) build(x->ch[1], x, mid + 1, r); 102 update(x); 103 } 104 105 void insert(int pos, char c) { 106 splay(kth(pos), nil); 107 splay(kth(pos + 1), root); 108 root->ch[1]->ch[0] = new_node(c, root->ch[1]); 109 update(root->ch[1]); update(root); 110 } 111 112 void modify(int pos, char c) { 113 splay(kth(pos), nil); 114 root->c = c; 115 update(root); 116 } 117 118 void remove(int pos) { 119 splay(kth(pos - 1), nil); 120 splay(kth(pos + 1), root); 121 del_node(root->ch[1]->ch[0]); 122 update(root->ch[1]); update(root); 123 } 124 125 bool check(int pos, int len) { 126 splay(kth(pos - len - 1), nil); 127 splay(kth(pos + len + 1), root); 128 splay(kth(pos), root->ch[1]); 129 Node* x = root->ch[1]->ch[0]; 130 return x->lhash == x->rhash; 131 } 132 133 int query(int pos) { 134 int l = 1, r = min(pos - 2, root->size - 1 - pos) + 1; 135 while(l < r) { 136 int mid = (l + r) >> 1; 137 if(check(pos, mid)) l = mid + 1; 138 else r = mid; 139 } 140 return 2 * l - 1; 141 } 142 143 void debug(Node* x) { 144 static int t = 0; 145 if(x == root) printf("Debug %d\n", ++t); 146 printf("val:%d lson:%d rson:%d lhash:%d rhash:%d\n", x - statePool, x->ch[0] - statePool, x->ch[1] - statePool, x->lhash, x->rhash); 147 if(x->ch[0] != nil) debug(x->ch[0]); 148 if(x->ch[1] != nil) debug(x->ch[1]); 149 } 150 } splay; 151 152 int main() { 153 scanf("%s", s + 1); 154 n = strlen(s + 1); 155 initBase(); 156 splay.init(); 157 splay.build(splay.root, splay.nil, 0, n + 1); 158 scanf("%d", &m); 159 char c; 160 for(int i = 0, a; i < m; ++i) { 161 scanf("%s%d", op, &a); 162 ++a; 163 if(strcmp(op, "Insert") == 0) { 164 scanf(" %c", &c); 165 splay.insert(a, c); 166 } 167 if(strcmp(op, "Delete") == 0) 168 splay.remove(a); 169 if(strcmp(op, "Update") == 0) { 170 scanf(" %c", &c); 171 splay.modify(a, c); 172 } 173 //splay.debug(splay.root); 174 if(strcmp(op, "Query") == 0) 175 printf("%d\n", splay.query(a)); 176 } 177 }
ACdream 1104 瑶瑶想找回文串(SplayTree + Hash + 二分),布布扣,bubuko.com
ACdream 1104 瑶瑶想找回文串(SplayTree + Hash + 二分)
标签:des style blog http color os io for
原文地址:http://www.cnblogs.com/oyking/p/3896588.html