Query
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2573 Accepted Submission(s): 851
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
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
题目意思：
给两个字符串s1, s2， 然后m组操作，共有两种操作1和2， 若为1，输入a, i, c即把第a个字符串（a==1||a==2）的第 i 个位置处的字符用 c 替换；若为2，输入x，即s1,s2都从x开始匹配最远到第j处位置，输出j。
思路：
把两个字符串比较相同的元素为1否则为0放在一个数组p[]中，然后把p[]挂在线段树叶子节点处，然后往上更新。
线段树节点包含这么几个元素：l, r(左边界、右边界), lm, rm(从左边界开始1的最长连续段的长度，以右边界为止1的最长连续段的长度)， mm(线段中1的最长的连续段长度)。（区间合并的老套路。。。）
其他的基本上就是老套路了，主要注意一下求和输出，还是直接看代码比较容易上手：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 using namespace std;
6 #define N 1000005
7 #define ll root<<1
8 #define rr root<<1|1
9 #define mid (a[root].l+a[root].r)/2
10
11 int p[1000005];
12 char s1[1000005];
13 char s2[1000005];
14
15 struct node{
16 int l, r;
17 int lm, rm, mm;
18 }a[4*N];
19
20
21 void up(int root){ //貌似线段树区间合并都有这个函数
22 if(a[root].l==a[root].r) return;
23 a[root].lm=a[ll].lm;
24 a[root].rm=a[rr].rm;
25 a[root].mm=max(a[ll].mm,a[rr].mm);
26 if(a[ll].lm==a[ll].r-a[ll].l+1) a[root].lm+=a[rr].lm;
27 if(a[rr].rm==a[rr].r-a[rr].l+1) a[root].rm+=a[ll].rm;
28 a[root].mm=max(max(a[root].lm,a[root].rm),max(a[root].mm,a[ll].rm+a[rr].lm));
29 }
30
31 void build(int l,int r,int root){
32 a[root].l=l;
33 a[root].r=r;
34 if(l==r){
35 a[root].lm=a[root].rm=a[root].mm=p[l];
36 return;
37 }
38 build(l,mid,ll);
39 build(mid+1,r,rr);
40 up(root);
41 }
42
43 void update(int q,int val,int root){
44 if(a[root].l==q&&a[root].r==q){
45 a[root].lm=a[root].mm=a[root].rm=val;
46 return;
47 }
48 if(q<=mid) update(q,val,ll);
49 else update(q,val,rr);
50 up(root);
51 }
52
53 int query(int q,int root){
54 if(a[root].l==a[root].r){
55 return a[root].mm;
56 }
57 if(q<=mid){
58 if(q>=a[ll].r-a[ll].rm+1) return a[ll].r-q+1+query(mid+1,rr); //若q在左子树且在左子树1右连续段则需要加上右子树1左连续段
59 else return query(q,ll);
60 }
61 else{
62 return query(q,rr);
63 }
64 }
65
66 main()
67 {
68 int t, kase=1;
69 int i, j, k, m;
70 int n1, n2;
71 int x, y, z;
72 char c[5];
73 scanf("%d",&t);
74 while(t--){
75 //memset(p,0,sizeof(p));
76 printf("Case %d:\n",kase++);
77 scanf("%s%s",s1,s2);
78 n1=strlen(s1);
79 n2=strlen(s2);
80 for(i=0;i<min(n1,n2);i++){
81 if(s1[i]==s2[i]) p[i]=1; //相同为1否则为0
82 else p[i]=0;
83 }
84 for(i=min(n1,n2);i<max(n1,n2);i++){
85 p[i]=0;
86 }
87 build(0,max(n1,n2)-1,1);
88 scanf("%d",&m);
89 while(m--){
90 scanf("%d",&z);
91 if(z==1){
92 scanf("%d %d %s",&x,&y,c);
93 if(x==1){
94 s1[y]=c[0];
95 }
96 else s2[y]=c[0];
97 p[y]=(s1[y]==s2[y]?1:0);
98 update(y,p[y],1);
99 }
100 else{
101 scanf("%d",&x);
102 printf("%d\n",query(x,1));
103 }
104 }
105 }
106 }
HDU 4339 线段树区间合并