HDU 4339 Query（线段树）

Problem Description

You are given two strings s1[0..l1], s2[0..l2] and Q - number of queries.
Your task is to answer next queries:
  1) 1 a i c - you should set i-th character in a-th string to c;
  2) 2 i - you should output the greatest j such that for all k (i<=k and k<i+j) s1[k] equals s2[k].

 

Input

The first line contains T - number of test cases (T<=25).
Next T blocks contain each test.
The first line of test contains s1.
The second line of test contains s2.
The third line of test contains Q.
Next Q lines of test contain each query:
  1) 1 a i c (a is 1 or 2, 0<=i, i<length of a-th string, 'a'<=c, c<='z')
  2) 2 i (0<=i, i<l1, i<l2)
All characters in strings are from 'a'..'z' (lowercase latin letters).
Q <= 100000.
l1, l2 <= 1000000.

 

Output

For each test output "Case t:" in a single line, where t is number of test (numbered from 1 to T).
Then for each query "2 i" output in single line one integer j.

 

Sample Input

1aaabbaaabbaa72 02 12 22 31 1 2 b2 02 3

 

Sample Output

Case 1:210141

 

Source

2012 Multi-University Training Contest 4

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 1000100;int tcase, q;int cc[maxn<<2][2];int llen[maxn<<2], rlen[maxn<<2], mlen[maxn<<2]; // 左端点开始的最大的相同长度，右端开始的最大的相同长度，整个区间里的最大相同长度char s1[maxn], s2[maxn];void pushUp(int l, int r, int rt){    int ls = rt << 1;    int rs = rt << 1 | 1;    int m = (l + r) >> 1;        llen[rt] = llen[ls];    rlen[rt] = rlen[rs];    if (llen[ls] == m - l + 1) {        llen[rt] += llen[rs];     }     if (rlen[rs] == r - m) {        rlen[rt] += rlen[ls];    }     mlen[rt] = max(rlen[ls] + llen[rs], max(llen[rt], rlen[rt]));}void build(int l, int r, int rt){    if (l == r) {        cc[rt][0] = s1[l];        cc[rt][1] = s2[l];        if (cc[rt][0] == cc[rt][1]) {            rlen[rt] = llen[rt] = mlen[rt] = 1;        } else {            rlen[rt] = llen[rt] = mlen[rt] = 0;        }        return ;    }     int m = (l + r) >> 1;    build(l, m, rt << 1);    build(m + 1, r, rt << 1 | 1);    pushUp(l, r, rt); }void update(int l, int r, int rt, int p, int id, int c){    if (l == r) {        cc[rt][id] = c;        if (cc[rt][id] == cc[rt][!id]) {            llen[rt] = rlen[rt] = mlen[rt] = 1;        } else {            llen[rt] = rlen[rt] = mlen[rt] = 0;        }        return ;    }    int m = (l + r) >> 1;    if (p <= m) {        update(l, m, rt << 1, p, id, c);    } else {        update(m + 1, r, rt << 1 | 1, p, id, c);    }    pushUp(l, r, rt);}int query(int l, int r, int rt, int from){    if (l == r) {        return mlen[rt];    }    int m = (l + r) >> 1;    if (from <= m) {        if (rlen[rt<<1] >= m - from + 1) {            return m - from + 1 + query(m + 1, r, rt << 1 | 1, m + 1);        } else {            return query(l, m, rt << 1, from);        }    } else {        return query(m + 1, r, rt << 1 | 1, from);    }}int main(){    scanf("%d", &tcase);    for (int cas = 1; cas <= tcase; ++cas) {        printf("Case %d:\n", cas);        scanf("%s", s1);        scanf("%s", s2);        int L = min(strlen(s1), strlen(s2));        build(0, L - 1, 1);        scanf("%d", &q);        int op;        for (int i = 0; i < q; ++i) {            scanf("%d", &op);            if (op == 1) {                int id, p;                char C;                scanf("%d %d %c", &id, &p, &C);                if (p >= L) {                    continue;                }                update(0, L - 1, 1, p, id - 1, C);            } else {                int from;                scanf("%d", &from);                if (from >= L) {                    printf("0\n");                    continue;                }                printf("%d\n", query(0, L - 1, 1, from));            }        }    }    return 0;}