Line tree-hdu-4339-Query

Question meaning:

Give two strings. There are two types of operations.

1. Change a character at a certain position of a string.

2. Find the maximum number of two consecutive identical characters at the beginning of a position.

Solution:

The line segment tree maintains two values: the leftmost continuous maximum value of the interval and the rightmost continuous maximum value.

Every question is made, so we can stop thinking and abstract a little bit of experience.

Code:

 

# Include <iostream> # include <cmath> # include <cstdio> # include <cstdlib> # include <string> # include <cstring> # include <algorithm> # include <vector> # include <map> # include <set> # include <stack> # include <list> # include <queue> # define eps 1e-6 # define INF 0x1f1f1f1f # define PI acos (-1.0) # define ll _ int64 # define lson l, m, (rt <1) # define rson m + 1, r, (rt <1) | 1 // # pragma comment (linker, "/STACK: 1024000000,1024000000") using names Pace std; # define Maxn 1100000/* freopen ("data. in "," r ", stdin); freopen (" data. out "," w ", stdout); */struct Node {int ls, rs;} node [Maxn * 4]; // The maintenance interval of the Line Segment tree, 1. including the maximum number of consecutive equal numbers at the left endpoint, // 2. the maximum number of consecutive equal numbers at the right endpoint. char sa1 [Maxn], sa2 [Maxn]; int nm, n; void pushup (int rt, int num) // update {node [rt] upwards. ls = node [rt <1]. ls; node [rt]. rs = node [rt <1 | 1]. rs; if (node [rt]. ls> = num-(num> 1) // The number of left child ranges is num-(num> 1) node [rt]. ls + = node [rt <1 | 1]. ls; if (Node [rt]. rs> = (num> 1) // The number of right child ranges is num> 1 node [rt]. rs + = node [rt <1]. rs;} void build (int l, int r, int rt) {if (l = r) {if (l <nm) // The minimum length of the two strings node [rt]. ls = node [rt]. rs = (sa1 [l] = sa2 [r])? 1:0; else node [rt]. ls = node [rt]. rs = 0; return;} int m = (l + r)> 1; build (lson); build (rson); pushup (rt, r-l + 1);} void update (int k, int l, int r, int rt) {if (l = r) {if (l <nm) node [rt]. ls = node [rt]. rs = (sa1 [l] = sa2 [r])? 1:0; else node [rt]. ls = node [rt]. rs = 0; return;} int m = (l + r)> 1; if (k <= m) update (k, lson); else update (k, rson); pushup (rt, r-l + 1);} int query (int k, int l, int r, int rt) {if (l = r) return 1; // The position must be the same int m = (l + r)> 1; if (k <= m) {if (node [rt <1]. rs> m-k) // return m-k + 1 + node [rt <1 | 1]. ls; else return query (k, lson);} else return query (k, rson);} int main () {// scanf ("% s", sa1 ); // scanf ("% s", sa1 );// Printf ("% c \ n", sa1 [5]); int t; int AB, po, m, ki; char s [5]; scanf ("% d ", & t); for (int ca = 1; ca <= t; ca ++) {scanf ("% s", sa1); scanf ("% s ", sa2); int len1 = strlen (sa1), len2 = strlen (sa2); n = max (len1, len2); nm = min (len1, len2); build (0, n-1, 1); printf ("Case % d: \ n", ca); scanf ("% d", & m); while (m --) {scanf ("% d", & ki); if (ki = 1) {scanf ("% d % s", & AB, & po, s ); if (AB = 1) sa1 [po] = * s; else sa2 [po] = * s; update (po, 0, n-1, 1);} els E {scanf ("% d", & po); if (po> = nm | sa1 [po]! = Sa2 [po]) {printf ("0 \ n"); continue;} printf ("% d \ n", query (po, 0, n-1, 1); }}return 0 ;}