Codeforce E. Lucky Queries 线段树实践 - c++编程基础

Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya brought home string s with the length of n. The string only consists of lucky digits. The digits are numbered from the left to the right starting with 1. Now Petya should execute m queries of the following form:

switch l r ― "switch" digits (i.e. replace them with their opposites) at all positions with indexes from l to r, inclusive: each digit 4 is replaced with 7 and each digit 7 is replaced with 4 (1 ≤ l ≤ r ≤ n);

count ― find and print on the screen the length of the longest non-decreasing subsequence of string s.

Subsequence of a string s is a string that can be obtained from s by removing zero or more of its elements. A string is called non-decreasing if each successive digit is not less than the previous one.

Help Petya process the requests.

Input

The first line contains two integers n and m (1 ≤ n ≤ 106, 1 ≤ m ≤ 3 105) ― the length of the string s and the number of queries correspondingly. The second line contains n lucky digits without spaces ― Petya's initial string. Next m lines contain queries in the form described in the statement.

Output

For each query count print an answer on a single line.

Sample test(s) input

2 3
47
count
switch 1 2
count

output

2
1

数据量好大，使用线段树查询可以大大提高速度。

本题关键难点是线段树的动态更新，需要设置一个Lazy标志，就是不需要每次更新所有点，而是等到下次需要查询到这个点的时候，更新这个点的两个孩子节点。

其中的逻辑关系，需要好好研究研究代码吧。没搞清其中的逻辑，还是十分有难度的。

线段树本题就两个点了：

1 二分法操作树

2 Lazy标志动态更新区间

#include 
    
     
#include 
     
       #include 
      
        #include 
       
         #include 
        
          using namespace std; class LuckyQueries { int n, tSize; char *str; struct Node { int L4, L7, L47, L74; bool lazy; }; Node *segTree; void updateNode(int id, int left, int right) { segTree[id].L4 = segTree[left].L4 + segTree[right].L4; segTree[id].L7 = segTree[left].L7 + segTree[right].L7; int a = segTree[left].L47 + segTree[right].L7; int b = segTree[left].L4 + segTree[right].L47; segTree[id].L47 = max(a, b); a = segTree[left].L74 + segTree[right].L4; b = segTree[left].L7 + segTree[right].L74; segTree[id].L74 = max(a, b); segTree[id].lazy = false; } void conHelper(int low, int up, int id = 0) { if (low == up) { if (str[low] == '4') { segTree[id].L4 = 1, segTree[id].L7 = 0; segTree[id].L47 = 1, segTree[id].L74 = 0; } else { segTree[id].L4 = 0, segTree[id].L7 = 1; segTree[id].L47 = 0, segTree[id].L74 = 1; } segTree[id].lazy = false; return ; } int mid = low + ((up-low)>>1); int left = id*2+1; int right = id*2+2; conHelper(low, mid, left); conHelper(mid+1, up, right); updateNode(id, left, right); } void conTree() { int h = (int) ceil((double)log(double(n))/log(2.0)) + 1; tSize = (int) pow(2.0, double(h)) - 1; segTree = (Node *) malloc(sizeof(Node) * tSize); conHelper(0, n-1); } inline int getLongestIncease() { return max(segTree[0].L4, max(segTree[0].L47, segTree[0].L7)); } void invert(int root) { segTree[root].lazy = !segTree[root].lazy; swap(segTree[root].L4, segTree[root].L7); swap(segTree[root].L47, segTree[root].L74); } void updateTree(const int low, const int up, int tLow, int tUp, int root = 0) { if (tUp < low || up < tLow)//错写||成&& { return ; } if (low <= tLow && tUp <= up) { invert(root); return ; } int tMid = tLow + ((tUp-tLow)>>1); int left = root * 2 + 1; int right = root * 2 + 2; if (segTree[root].lazy) { segTree[root].lazy = false; invert(left); invert(right); } updateTree(low, up, tLow, tMid, left); updateTree(low, up, tMid+1, tUp, right); updateNode(root, left, right); } public: LuckyQueries() { int m; scanf("%d %d", &n, &m); getchar();//别忘记了去掉一个换行符 st

Codeforce E. Lucky Queries 线段树实践(一)