标签：
De Prezer loves palindrome strings. A string s1s2...sn is palindrome if and only if it is equal to its reverse.
De Prezer also loves queries.
You are given string s of length n and m queries. There are 3 types of queries :
1. 1px : Modify sp = x where 1 ≤ p ≤ n and x is a lower case English letter.
2. 2p : Print the length of the largest palindrome substring of s like slsl + 1...sr such that l ≤ p ≤ r and r - p = p - l. (1 ≤ p ≤ n)
3. 3p : Print the length of the largest palindrome substring of s like slsl + 1...sr such that l ≤ p and p + 1 ≤ r and r - p - 1 = p - l. (1 ≤ p ≤ n - 1) or - 1 if there is no such substring.
The first line of input contains s and m.
Next m lines contain queries.
1 ≤ n, m ≤ 105
s only contains lower case English letters.
For each query of type 2 and 3 print the answer in a single line.
解题报告：
大意就是给你一个字符串，有三种操作:
操作1：将某个位置的字符改成另外一个字符
操作2：询问以位置 x 位中点的奇回文串长度
操作3：询问以位置 x,x+1位中的偶回文串长度
我们建立一颗线段树来解决这个问题
线段树中的存储的 key1 表示 s[L] * p^0 + S[L+1] * p ^1 + ..... S[R] * p ^ (R-L) , key2 值表示S[R] * p ^ 0 + S[R-1] * p ^1 + ...S[L] * p^(R-L)
显然key2 是倒着的，这样方便我们查询这个区间是否是回文子串
题目显然要求是单点更新，区间查询.
对于单点更新而言，最重要的就是push_up函数，我们考虑
将[L,mid] 和 [mid + 1 , R]进行合并时，注意到key值的含义，我们需要给右边的[mid+1,R]的key1值乘上p^(mid-L+1),key2则是给[L,mid]乘上(R-mid+1)
那么更新我们就解决了,查询呢
下面皆以 key1 为例(key2同理可得)
对于key1而言，我们主要就是要给落在右边的查询区间段上的乘上一个值.
假设我们现在要查询的区间段是[ ql , qr ] ,此时所在的线段是[L,R],中点是mid , 我们假设 qr > mid(即有落在右边的部分)
那么我们需要给右边乘的值就是p ^ (mid - ql + 1) <仔细想想>
key2同理可得，这里不再累述
这样我们就解决了查询问题，那么对于每个询问，我们二分答案即可
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <cmath> #include <vector> #include <stack> #include <map> #include <set> #include <queue> #include <iomanip> #include <string> #include <ctime> typedef unsigned char byte; #define pb push_back #define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0) #define local freopen("in.txt","r",stdin) #define pi acos(-1) using namespace std; typedef pair<unsigned int ,unsigned int> dl; const int maxn = 1e5 + 500; const unsigned int p1 = 131; const unsigned int p2 = 171; unsigned int value1[maxn]; unsigned int value2[maxn]; typedef struct treenode { int l , r ; dl key1,key2; void updata(unsigned int v) { key1.first = key1.second = v; key2.first = key2.second = v; } }; treenode tree[maxn * 4]; inline void build_tree(int o,int l,int r) { tree[o].l = l , tree[o].r = r , tree[o].key1.first = tree[o].key1.second = tree[o].key2.first = tree[o].key2.second = 0; if (r > l) { int mid = l + (r-l)/2; build_tree(2*o,l,mid); build_tree(2*o+1,mid+1,r); } } inline void push_up(int o) { int ac = tree[2*o].r - tree[2*o].l + 1 ; int ac2 = tree[2*o+1].r - tree[2*o+1].l + 1; tree[o].key1.first = tree[2*o].key1.first + tree[2*o+1].key1.first * value1[ac]; tree[o].key1.second = tree[2*o].key1.second + tree[2*o+1].key1.second * value2[ac]; tree[o].key2.first = tree[2*o].key2.first * value1[ac2] + tree[2*o+1].key2.first; tree[o].key2.second = tree[2*o].key2.second * value2[ac2] + tree[2*o+1].key2.second; } void updata(int ql,int qr,int o,unsigned int v) { int l = tree[o].l , r = tree[o].r; if (ql <= l && qr >= r) tree[o].updata(v); else { int mid = l + (r-l) / 2; if (mid >= ql) updata(ql,qr,2*o,v); if (mid < qr) updata(ql,qr,2*o+1,v); push_up(o); } } dl query(int ql,int qr,int o,int type) { int l = tree[o].l , r = tree[o].r; //cout << "L is " << l << " R is " << r << " ql is " << ql << " qr is " << qr << endl; if (ql == l && qr == r) { if (type == 1) return tree[o].key1; else return tree[o].key2; } else { int mid = l + (r-l) / 2 , ac; dl res(0,0) , temp; if (mid >= ql) { int atr = min(mid,qr); temp = query(ql,atr,2*o,type); if (type == 2 && qr - mid > 0) { temp.first *= value1[qr - mid]; temp.second *= value2[qr - mid]; } res.first += temp.first; res.second += temp.second; } if (mid < qr) { int ltr = max(ql,mid+1); temp = query(ltr,qr,2*o+1,type); if (type == 1 && mid - ql + 1 > 0) { temp.first *= value1[mid - ql + 1]; temp.second *= value2[mid - ql + 1]; } res.first += temp.first; res.second += temp.second; } //cout << "L is " << l << " R is " << r << " fist-value is " << res.first << " second-value is " << res.second << endl; return res; } } char str[maxn] , temp[200]; int length , m ; bool equaldl(dl x,dl y) { return x.first == y.first && x.second == y.second; } int main(int argc,char *argv[]) { scanf("%s%d",str,&m); value1[0] = value2[0] = 1; length = strlen(str); build_tree(1,0,length-1); for(int i = 1 ; i <= length ; ++ i) { value1[i] = value1[i-1]*p1; value2[i] = value2[i-1]*p2; } for(int i = 0 ; i < length ; ++ i) updata(i,i,1,(unsigned int)str[i]); while(m--) { int type,x; scanf("%d",&type); if (type == 1) { scanf("%d%s",&x,temp); x--; str[x] = temp[0]; updata(x,x,1,temp[0]); } else { scanf("%d",&x); int L , R; x--; if (type == 2) { L = 1 , R = min( x + 1 , length - x ); while(L < R) { int mid = L + (R-L+1) / 2; int tl = x - mid + 1 ; int tr = x + mid - 1 ; if (equaldl(query(tl,tr,1,1),query(tl,tr,1,2))) L = mid; else R = mid - 1; } printf("%d\n",2*L-1); } else { if (x+1 >= length || str[x] != str[x+1]) printf("-1\n"); else { int L = 1 , R = min(x + 1 , length - x - 1); while(L < R) { int mid = L + (R-L+1) / 2; int tl = x - mid + 1; int tr = x + mid; if (equaldl(query(tl,tr,1,1),query(tl,tr,1,2))) L = mid; else R = mid - 1; } printf("%d\n",2*L); } } } } return 0; }
Gym 100570E : Palindrome Query
标签：
原文地址：http://www.cnblogs.com/Xiper/p/4695435.html
