哈希
一维哈希
纯模板
const ull B = 1e8 + 7;
const int INF = 0x3f3f3f3f3f3f;
// a在b中
bool contain(string a, string b) {
int a1 = a.length();
int b1 = b.length();
if(a1 > b1) return 0;
ull t = 1;
for(int i = 0; i < a1; i++) t *= B;
ull ah = 0, bh = 0;
for(int i = 0; i < a1; i++) ah = ah * B + a[i];
for(int i = 0; i < a1; i++) bh = bh * B + b[i];
for(int i = 0; i + a1 <= b1; i++) {
if(ah == bh) return true;
if(i + a1 < b1) bh = bh * B + b[i + a1] - b[i] * t;
}
return false;
}
//a 的后缀等于 b 的前缀
int overlap(string a, string b) {
int a1 = a.length();
int b1 = b.length();
ull ah = 0, bh = 0, t = 1;
int ans = 0;
for(int i = 1; i <= min(a1, b1); i++) {
ah = ah + a[a1 - i] * t;
bh = bh * B + b[i - 1];
if(ah == bh) ans = i;
t *= B;
}
return ans;
}
二维哈希
例题
POJ 3690.
题意：
给一个nm的矩阵，再给T个pq的小矩阵，问有多少个小矩阵在大矩阵中出现了.
输入：
输入包含多个测试用例。
每个测试用例均以包含五个整数N，M，T，P和Q的行开始（1≤N，M≤1000，1≤T≤100，1≤P，Q≤50）。
接下来的N行描述了N×M矩阵，每个矩阵包含M个字符“ *”或“ 0”。
测试用例的最后一部分描述了T个星座，每个星座采用与描述天空的矩阵相同格式的P条线。
每个星座前面都有一个空白行。
最后一个测试用例后面是包含五个零的行
输出：
对于每个测试用例，请打印一行，其中包含测试用例编号（以1开头），后跟天空中出现的星座数量。
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<vector>
#include<cstdio>
#include<set>
#define ull unsigned long long
using namespace std;
int N, M, T, P, Q;
const int MAX = 1e3 + 10;
char field[MAX][MAX];
char patterns[MAX][MAX];
ull hash[MAX][MAX], temp[MAX][MAX];
void compute_hash(char a[MAX][MAX], int n, int m) {
const ull B1 = 1e9 - 7;
const ull B2 = 1e9 + 7;
ull t1 = 1;
for(int j = 0; j < Q; j++) t1 *= B1;
for(int i = 0; i < n; i++) {
ull e = 0;
for(int j = 0; j < Q; j++) e = e * B1 + a[i][j];
for(int j = 0; j + Q <= m; j++) {
temp[i][j] = e;
if(j + Q < m) e = e * B1 + a[i][j + Q] - a[i][j] * t1;
}
}
ull t2 = 1;
for(int i = 0; i < P; i++) t2 *= B2;
for(int j = 0; j + Q <= m; j++) {
ull e = 0;
for(int i = 0; i < P; i++) e = e * B2 + temp[i][j];
for(int i = 0; i + P <= n; i++) {
hash[i][j] = e;
if(i + P < n) e = e * B2 + temp[i + P][j] - temp[i][j] * t2;
}
}
}
int main() {
multiset<ull> unseen;
int cnt = 1;
while(scanf("%d%d%d%d%d", &N, &M, &T, &P, &Q) != EOF && N, M, P, T, Q) {
unseen.clear();
for(int i = 0; i < N; i++)
scanf("%s", field[i]);
int tt = T;
while(tt--) {
for(int i = 0; i < P; i++)
scanf("%s", patterns[i]);
compute_hash(patterns, P, Q);
unseen.insert(hash[0][0]);
}
compute_hash(field, N, M);
for(int i = 0; i + P <= N; i++) {
for(int j = 0; j + Q <= M; j++)
unseen.erase(hash[i][j]);
}
//cout << T << " " << unseen.size() << endl;
int ans = T - unseen.size();
printf("Case %d: %d\n", cnt, ans);
cnt++;
}
}
后缀数组
纯模板
const int MAX = 2e5 + 10;
int rank[MAX];
int n, k;
bool compare_sa(int i, int j) {
if(rank[i] != rank[j]) return rank[i] < rank[j];
else {
int A = i + k <= n ? rank[i + k] : -1;
int B = j + k <= n ? rank[j + k] : -1;
return A < B;
}
}
int temp[MAX];
void construct_sa(int *S, int n, int *sa) {
for(int i = 0; i <= n; i++) {
sa[i] = i;
rank[i] = i < n ? S[i] : -1;
}
for(k = 1; k <= n; k <<= 1) {
sort(sa, sa + n + 1, compare_sa);
temp[sa[0]] = 0;
for(int i = 1; i <= n; i++) {
temp[sa[i]] = temp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
}
for(int i = 0; i <= n; i++) {
rank[i] = temp[i];
}
}
}
实例 POJ 3581
题意：
给一串数，将其分成三个区间并且颠倒这三个区间，使得新数列字典序最小
输入：
n个，接下来n行一行一个
输出：
n个，一行一个
#include <cstdio>
#include <cstring>
#include <string>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
const int MAX = 2e5 + 10;
int rank[MAX];
int n, k;
bool compare_sa(int i, int j) {
if(rank[i] != rank[j]) return rank[i] < rank[j];
else {
int A = i + k <= n ? rank[i + k] : -1;
int B = j + k <= n ? rank[j + k] : -1;
return A < B;
}
}
int temp[MAX];
void construct_sa(int *S, int n, int *sa) {
for(int i = 0; i <= n; i++) {
sa[i] = i;
rank[i] = i < n ? S[i] : -1;
}
for(k = 1; k <= n; k <<= 1) {
sort(sa, sa + n + 1, compare_sa);
temp[sa[0]] = 0;
for(int i = 1; i <= n; i++) {
temp[sa[i]] = temp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
}
for(int i = 0; i <= n; i++) {
rank[i] = temp[i];
}
}
}
int a[MAX];
int rev[MAX * 2], sa[MAX * 2];
void solve() {
reverse_copy(a, a + n, rev);
construct_sa(rev, n, sa);
int p1;
for(int i = 0; i <= n; i++) {
p1 = n - sa[i];
if(p1 >= 1 && sa[i] >= 2)
break;
}
int m = n - p1;
reverse_copy(a + p1, a + n, rev);
reverse_copy(a + p1, a + n, rev + m);
construct_sa(rev, 2 * m, sa);
int p2;
for(int i = 0; i <= 2 * m; i++) {
p2 = n - sa[i];
if(p2 + m - n >= 1 && sa[i] >= 1)
break;
}
reverse(a, a + p1);
reverse(a + p1, a + p2);
reverse(a + p2, a + n);
for(int i = 0; i < n; i++) {
printf("%d\n", a[i]);
}
}
int main() {
scanf("%d", &n);
for(int i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
solve();
}
编辑距离
模板
/**
dp[i][j]：所有将a中的前i个字母变成 b中前j个字母的集合的操作次数
对a中的第i个字母操作，分为以下四种操作
1. 在该字母之后添加: dp[i][j] = dp[i][j-1] + 1
2. 删除该字母: dp[i][j] = dp[i-1][j] + 1
3. 替换该字母: dp[i][j] = dp[i-1][j-1] + 1
4. 啥也不做(对应结尾字母相同): dp[i][j] = dp[i-1][j-1]
**/
int edit_distance(char s1[], char s2[], int n, int m){
for (int i = 1; i <= n; i++) dp[i][0] = i;
for (int i = 1; i <= m; i++) dp[0][i] = i;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++){
if (s1[i] == s2[j]) dp[i][j] = dp[i - 1][j - 1];
else dp[i][j] = min(min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
}
return dp[n][m];
}
题意：
编辑距离，是指两个字串之间，由一个转成另一个所需的最少编辑操作次数。许可的编辑操作包括将一个字符替换成另一个字符，插入一个字符，删除一个字符。
输入：
两行，两个字符串a,b
输出：
求a和b的编辑距离
//下面就是学长的代码
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
char s1[N], s2[N];
int dp[N][N];
int edit_distance(char s1[], char s2[], int n, int m){
for (int i = 1; i <= n; i++) dp[i][0] = i;
for (int i = 1; i <= m; i++) dp[0][i] = i;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= m; j++){
if (s1[i] == s2[j]) dp[i][j] = dp[i - 1][j - 1];
else dp[i][j] = min(min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
}
return dp[n][m];
}
int main(){
cin >> s1 + 1;
cin >> s2 + 1;
int n = strlen(s1 + 1);
int m = strlen(s2 + 1);
cout << edit_distance(s1, s2, n, m);
}
manacher
纯模板
string manacher(string s) {
//预处理
string t = "$#";
for(int i = 0; i < s.size(); i++) {
t += s[i];
t += '#';
}
vector <int> p(t.size(), 0);
int maxx = 0, id = 0;
int center = 0, len = 0;
for(int i = 0; i < t.size(); i++) {
p[i] = maxx > i ? min(p[2 * id - i], maxx - i) : 1;
while(t[i + p[i]] == t[i - p[i]]) p[i]++;
if(i + p[i] > maxx) {
maxx = i + p[i];
id = i;
}
if(p[i] > len) {
len = p[i];
center = i;
}
}
//center（结果最大回文串中心下标）,len（最大长回文长度)
return s.substr((center - Len) / 2 , len - 1);
}
KMP
int nextt[MAXM];
int A[MAXN], B[MAXM];
int N, M;
int KMP() {
nextt[0] = -1;
int k = -1;//前缀
//j 后缀
for(int j = 0; j < M - 1;) {
if(k == -1 || B[j] == B[k]) {
j++;
k++;
nextt[j] = k;
}
else
k = nextt[k];
}
int i = 0, j = 0;
while(i < N && j < M) {
if(j == -1 || A[i] == B[j]) {
i++;
j++;
}
else j = nextt[j];
}
if(j == M) return i - j + 1;
else return -1;
}
tire 树
int tree[MAX][27];
int num[MAX];
int pos = 1;
void insert(char* word){
int c = 0;
for(int i = 0; word[i]; i++) {
int n = word[i] - 'a';
if(tree[c][n] == 0) tree[c][n] = pos++;
c = tree[c][n];
num[c]++;
}
}
int find(char* word) {
int c = 0;
for(int i = 0; word[i]; i++) {
int n = word[i] - 'a';
if(tree[c][n] == 0) return 0;
c = tree[c][n];
}
return num[c];
}
AC自动机
模板1，求个数
//从下标1开始
int N;
namespace AC {
int tot, tr[MAX_M][26];
int fail[MAX_N];
int idx[MAX_M];
void init() {
memset(fail, 0, sizeof(fail));
memset(tr, 0, sizeof(tr));
memset(idx, 0, sizeof(idx));
tot = 0;
}
void insert(char *s) {
int u = 0;
for(int i = 1; s[i]; i++) {
if(!tr[u][s[i] - 'a']) tr[u][s[i] - 'a'] = ++tot;
u = tr[u][s[i] - 'a'];
}
idx[u]++;
}
queue<int>q;
void build() {
for(int i = 0; i < 26; i++) {
if(tr[0][i]) q.push(tr[0][i]);
}
while(q.size()) {
int u = q.front();
q.pop();
for(int i = 0; i < 26; i++) {
if(tr[u][i]) {
fail[tr[u][i]] = tr[fail[u]][i];
q.push(tr[u][i]);
}
else
tr[u][i] = tr[fail[u]][i];
}
}
}
int query(char *t) {
int u = 0, res = 0;
for(int i = 1; t[i]; i++) {
u = tr[u][t[i] - 'a'];
for(int j = u; j && idx[j] != -1; j = fail[j]) {
res += idx[j];
idx[j] = -1;
}
}
return res;
}
}
模板1，数组形式
const int MAX_N = 1e4 + 10;
const int MAX_M = 1e6 + 10;
//从下标1开始
const int sizee = MAX_N * 80;
namespace AC {
int tot, tr[sizee][26];
int fail[sizee];
int idx[sizee];
void init() {
memset(fail, 0, sizeof(fail));
memset(tr, 0, sizeof(tr));
memset(idx, 0, sizeof(idx));
tot = 0;
}
void insert(char *s) {
int u = 0;
for(int i = 1; s[i]; i++) {
if(!tr[u][s[i] - 'a']) tr[u][s[i] - 'a'] = ++tot;
u = tr[u][s[i] - 'a'];
}
idx[u]++;
}
void build() {
int L = 0, R = 0,
q[MAX_M];
for(int i = 0; i < 26; i++) {
if(tr[0][i]) q[R++] = tr[0][i];
}
while(L < R) {
int u = q[L++];
for(int i = 0; i < 26; i++) {
if(tr[u][i]) {
fail[tr[u][i]] = tr[fail[u]][i];
q[R++] = tr[u][i];
}
else
tr[u][i] = tr[fail[u]][i];
}
}
}
int query(char *t) {
int u = 0, res = 0;
for(int i = 1; t[i]; i++) {
u = tr[u][t[i] - 'a'];
for(int j = u; j && idx[j] != -1; j = fail[j]) {
res += idx[j];
idx[j] = -1;
}
}
return res;
}
}
模板2，求个数等于个数最大的 个数
namespace AC {
const int sizee = N * 80;
int tot, tr[sizee][26];
int fail[sizee], idx[sizee], val[sizee];
int cnt[N];//第i个字符出现的次数
void init() {
memset(fail, 0, sizeof(fail));
memset(tr, 0, sizeof(tr));
memset(idx, 0, sizeof(idx));
memset(val, 0, sizeof(val));
memset(cnt, 0, sizeof(cnt));
tot = 0;
}
void insert(char *s, int id) {
int u = 0;
for(int i = 1; s[i]; i++) {
if(!tr[u][s[i] - 'a']) tr[u][s[i] - 'a'] = ++tot;
u = tr[u][s[i] - 'a'];
}
idx[u] = id;
}
queue<int>q;
void build() {
for(int i = 0; i < 26; i++) {
if(tr[0][i]) q.push(tr[0][i]);
}
while(q.size()) {
int u = q.front();
q.pop();
for(int i = 0; i < 26; i++) {
if(tr[u][i]) {
fail[tr[u][i]] = tr[fail[u]][i];
q.push(tr[u][i]);
}
else
tr[u][i] = tr[fail[u]][i];
}
}
}
int query(char *t) { // 返回最大的出现次数
int u = 0, res = 0;
for(int i = 1; t[i]; i++) {
u = tr[u][t[i] - 'a'];
for(int j = u; j; j = fail[j])
val[j]++;
}
for(int i = 0; i <= tot; i++) {
if(idx[i]) {
res = max(res, val[i]);
cnt[idx[i]] = val[i];
}
}
return res;
}
}
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