hdu 4339 Query
Query
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1009 Accepted Submission(s): 302
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].
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.
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.
Then for each query "2 i" output in single line one integer j.
Sample Input
1
aaabba
aabbaa
7
2 0
2 1
2 2
2 3
1 1 2 b
2 0
2 3
Sample Output
Case 1:
2
1
0
1
4
1
Source
Recommend
zhoujiaqi2010
//len代表区间最左开始最长公共子串
//然后 if(len[k<<1]==m-l+1) len[k]=len[k<<1]+len[k<<1|1]; else len[k]=len[k<<1];
//开心的今早上1Y，本来打算昨晚写的、结果去看爱情公寓3了、、^_^
#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #include <vector> #define N 1000000 #define lson l,m,k<<1 #define rson m+1,r,k<<1|1 using namespace std; char s1[N+2],s2[N+2]; int len[N<<2]; void up(int &k,int &l,int &m) { if(len[k<<1]==m-l+1) len[k]=len[k<<1]+len[k<<1|1]; else len[k]=len[k<<1]; } void build(int l,int r,int k) { if(l==r) { if(s1[l]==s2[l]) len[k]=1; else len[k]=0; return; } int m=(l+r)>>1; build(lson); build(rson); up(k,l,m); } void update(int &i,int l,int r,int k) { if(l==r) { if(s1[l]==s2[l]) len[k]=1; else len[k]=0; return; } int m=(l+r)>>1; if(i<=m) update(i,lson); else update(i,rson); up(k,l,m); } int query(int &i,int l,int r,int k) { if(l==r) { return len[k];} int m=(l+r)>>1,t; if(i<=m) { t=query(i,lson); if(t==m-i+1) return t+len[k<<1|1]; return t; } return query(i,rson); } int main() { int l,l1,l2; int t=1,T; int q; scanf("%d",&T); int op,a,i; char c; while(T--) { scanf("%s",s1); scanf("%s",s2); l1=strlen(s1); l2=strlen(s2); l=l1<l2?l1:l2;//选取短的串建树 l--; //因为下标从0开始 build(0,l,1); scanf("%d",&q); printf("Case %d:\n",t++); while(q--) { scanf("%d",&op); if(op==1) { scanf("%d %d %c",&a,&i,&c); if(i>l) continue;//超过长度更新没意义 if(a==1)s1[i]=c; else s2[i]=c; update(i,0,l,1); } else { scanf("%d",&i); if(i>l){printf("0\n");continue;}//超长的话肯定是0 printf("%d\n",query(i,0,l,1)); } } } return 0; }