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题意：给两串字符串，操作1：替换其中一个字符串的某个位置的字符 操作2：查询从p开始相等的最长连续长度
分析：树状数组可以维护一个区间内公共长度(连续)的情况，查询时用二分查找最远的端点即可。还可以用线段树去做，线段树能处理的问题很多，这题只要往右区间合并就行了。
收获：1.线段树的区间合并又做一题(虽然我写的还没AC) 2. 树状数组写起来方便又奇效！
代码1(树状数组)：
/************************************************
* Author :Running_Time
* Created Time :2015-8-26 9:46:09
* File Name :I.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e6 + 10;
const int Q = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
char s1[N], s2[N];
int L;
struct BIT {
int c[N];
void init(void) {
memset (c, 0, sizeof (c));
}
void updata(int i, int x) {
while (i <= L) {
c[i] += x; i += (i & (-i));
}
}
int query(int i) {
int ret = 0;
while (i) {
ret += c[i]; i -= (i & (-i));
}
return ret;
}
int bsearch(int p, int l, int r) {
while (l <= r) {
int mid = (l + r) >> 1;
if (query (mid) - query (p-1) == mid - p + 1) l = mid + 1;
else r = mid - 1;
}
return l - p;
}
}bit;
int main(void) {
int T, cas = 0; scanf ("%d", &T);
while (T--) {
scanf ("%s%s", s1 + 1, s2 + 1);
int q; scanf ("%d", &q);
int len1 = strlen (s1 + 1), len2 = strlen (s2 + 1);
L = min (len1, len2);
bit.init ();
for (int i=1; i<=L; ++i) {
if (s1[i] == s2[i]) bit.updata (i, 1);
}
printf ("Case %d:\n", ++cas);
while (q--) {
int op; scanf ("%d", &op);
if (op == 1) {
int x, p; char c; scanf ("%d%d %c", &x, &p, &c);
++p;
if (p > L) continue;
bool same = (s1[p] == s2[p]);
if (x == 1) {
s1[p] = c;
if (same && s1[p] != s2[p]) bit.updata (p, -1);
else if (!same && s1[p] == s2[p]) bit.updata (p, 1);
}
else {
s2[p] = c;
if (same && s1[p] != s2[p]) bit.updata (p, -1);
else if (!same && s1[p] == s2[p]) bit.updata (p, 1);
}
}
else {
int p; scanf ("%d", &p);
++p;
if (p > L) {
puts ("0"); continue;
}
printf ("%d\n", bit.bsearch (p, p, L));
}
}
}
return 0;
}
代码2(线段树)：
/************************************************
* Author :Running_Time
* Created Time :2015-8-26 9:46:09
* File Name :I.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e6 + 10;
const int Q = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
char s1[N], s2[N];
struct ST {
int sum[N<<2];
void push_up(int rt, int len) { //sum[rt<<1]表示从左子树线段左端点开始最长连续的公共的长度，如果全是公共的，
if (sum[rt<<1] == len - (len >> 1)) sum[rt] = sum[rt<<1] + sum[rt<<1|1]; //合并右子树的线段长度
else sum[rt] = sum[rt<<1];
}
void build(int l, int r, int rt) {
if (l == r) {
sum[rt] = s1[l] == s2[r] ? 1 : 0;
return ;
}
int mid = (l + r) >> 1;
build (lson); build (rson);
push_up (rt, r - l + 1);
}
void updata(int p, int c, int l, int r, int rt) {
if (l == r) {
sum[rt] = c; return ;
}
int mid = (l + r) >> 1;
if (p <= mid) updata (p, c, lson);
else updata (p, c, rson);
push_up (rt, r - l + 1);
}
int query(int p, int l, int r, int rt) {
if (l == r) return sum[rt];
int mid = (l + r) >> 1;
if (p <= mid) {
int ret = query (p, lson);
if (p + ret - 1 == mid) return ret + sum[rt<<1|1];
else return ret;
}
else return query (p, rson);
}
}st;
int main(void) {
int T, cas = 0; scanf ("%d", &T);
while (T--) {
scanf ("%s%s", s1 + 1, s2 + 1);
int q; scanf ("%d", &q);
int len1 = strlen (s1 + 1), len2 = strlen (s2 + 1);
int mn = min (len1, len2);
st.build (1, mn, 1);
printf ("Case %d:\n", ++cas);
while (q--) {
int op; scanf ("%d", &op);
if (op == 1) {
int x, p; char c; scanf ("%d%d %c", &x, &p, &c);
++p;
if (p > mn) continue; //特殊情况特判
if (x == 1) {
s1[p] = c;
}
else {
s2[p] = c;
}
st.updata (p, (s1[p] == s2[p]), 1, mn, 1);
}
else {
int p; scanf ("%d", &p);
++p;
if (p > mn) { //特殊情况特判
puts ("0"); continue;
}
printf ("%d\n", st.query (p, 1, mn, 1));
}
}
}
return 0;
}
代码3(unAC)：
/************************************************
* Author :Running_Time
* Created Time :2015-8-26 9:46:09
* File Name :I.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e6 + 10;
const int Q = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
char s1[N], s2[N];
struct ST {
int sum[N<<2], lsum[N<<2], rsum[N<<2];
void push_up(int rt, int len) {
lsum[rt] = lsum[rt<<1];
rsum[rt] = rsum[rt<<1|1];
sum[rt] = max (sum[rt<<1], sum[rt<<1|1]);
if (lsum[rt] == len - (len >> 1)) lsum[rt] += lsum[rt<<1|1];
if (rsum[rt] == len >> 1) rsum[rt] += rsum[rt<<1];
sum[rt] = max (rsum[rt<<1] + lsum[rt<<1|1], max (sum[rt<<1], sum[rt<<1|1]));
}
void build(int l, int r, int rt) {
if (l == r) {
sum[rt] = lsum[rt] = rsum[rt] = s1[l] == s2[r] ? 1 : 0;
return ;
}
int mid = (l + r) >> 1;
build (lson); build (rson);
push_up (rt, r - l + 1);
}
void updata(int p, int c, int l, int r, int rt) {
if (l == r) {
sum[rt] = lsum[rt] = rsum[rt] = c; return ;
}
int mid = (l + r) >> 1;
if (p <= mid) updata (p, c, lson);
else updata (p, c, rson);
push_up (rt, r - l + 1);
}
int query(int p, int l, int r, int rt) {
if (l == r) return sum[rt];
int mid = (l + r) >> 1;
if (p <= mid) {
if (p + rsum[rt<<1] - 1 >= mid) return mid - p + 1 + query (p, rson);
else return query (p, lson);
}
else return query (p, lson);
}
}st;
int main(void) {
int T, cas = 0; scanf ("%d", &T);
while (T--) {
scanf ("%s%s", s1 + 1, s2 + 1);
int q; scanf ("%d", &q);
int len1 = strlen (s1 + 1), len2 = strlen (s2 + 1);
int mn = min (len1, len2);
st.build (1, mn, 1);
printf ("Case %d:\n", ++cas);
while (q--) {
int op; scanf ("%d", &op);
if (op == 1) {
int x, p; char c; scanf ("%d%d %c", &x, &p, &c);
if (p + 1 > mn) continue;
if (x == 1) {
s1[p+1] = c;
}
else {
s2[p+1] = c;
}
st.updata (p+1, (s1[p+1] == s2[p+1]), 1, mn, 1);
}
else {
int p; scanf ("%d", &p);
if (p + 1 > mn) {
puts ("0"); continue;
}
printf ("%d\n", st.query (p + 1, 1, mn, 1));
}
}
}
return 0;
}
二分+树状数组/线段树(区间更新) HDOJ 4339 Query
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原文地址：http://www.cnblogs.com/Running-Time/p/4760643.html
