HDU 4300 Clairewd’s message （kmp | exkmp）

题意:

给定明文→密文的转换表,给出密文明文的拼接字符串,明文缺失部分或者缺失全部,还原最短的密文明文拼接字符串

分析:

明文缺失,明文肯定小于等于floor(总长度/2),太长了整个字符串就非法了

得到这个之后,我们取到后半部分长度为floor(总长度/2)的“明文”作为母串s

同时我们“翻译”原串由密文明文→明文XX(XX表示未知),作为t

然后我们匹配两段的最大重叠部分,就是现有的存在的明文长度了

匹配部分可以用kmp,问题得以解决

代码:

//
//  Created by TaoSama on 2015-10-30
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, nxt
;
char a[26], s
, t
;

void getNxt(char *t) {
nxt[0] = -1;
int i = 0, j = -1;
while(i < m) {
if(j == -1 || t[i] == t[j]) nxt[++i] = ++j;
else j = nxt[j];
}
}

int kmp(char *s, char *t) {
int i = 0, j = 0;
while(i < n) {
if(j == -1 || s[i] == t[j]) ++i, ++j;
else j = nxt[j];
}
return j; //s后缀与t前缀的最大重合长度
}

//s: 密明->明
//t: decode->明X
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int T; scanf("%d", &T);
while(T--) {
scanf("%s%s", a, s);
char decode[26];
for(int i = 0; i < 26; ++i) decode[a[i] - 'a'] = i + 'a';
for(int i = 0; m = i, s[i]; ++i) t[i] = decode[s[i] - 'a'];
t[m] = 0;
n = m >> 1;
getNxt(t);
int ans = m - kmp(s + m - n, t);
if(ans) s[ans] = t[ans] = 0;
printf("%s%s\n", s, t);
}
return 0;
}

分析:

用exkmp可以,前面的思路不变,求现有的明文长度的时候

枚举s的ex[i]就是s[i⋯n−1]与t的lcp,判断是否等于余下的串的长度,也是我们的所求

代码:

//
//  Created by TaoSama on 2015-10-30
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, nxt
, ex
;
char a[26], s
, t
;

void getNxt(char *t) {
nxt[0] = m;
int j = 0;
while(t[j + 1] && t[j + 1] == t[j]) ++j;
nxt[1] = j;

int a = 1, p, L;
for(int i = 2; i < m; ++i) {
p = nxt[a] + a - 1, L = nxt[i - a];
if(i - 1 + L < p) nxt[i] = L;
else {
j = max(0, p - i + 1);
while(t[i + j] && t[i + j] == t[j]) ++j;
nxt[i] = j;
a = i;
}
}
}

void getEx(char *s, char * t) {
int j = 0;
while(s[j] && t[j] && s[j] == t[j]) ++j;
ex[0] = j;

int a = 0, p, L;
for(int i = 1; i < n; ++i) {
p = ex[a] + a - 1, L = nxt[i - a];
if(i - 1 + L < p) ex[i] = L;
else {
j = max(0, p - i + 1);
while(s[i + j] && t[j] && s[i + j] == t[j]) ++j;
ex[i] = j;
a = i;
}
}
}

//s: 密明->明
//t: decode->明X
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int T; scanf("%d", &T);
while(T--) {
scanf("%s%s", a, s);
char decode[26];
for(int i = 0; i < 26; ++i) decode[a[i] - 'a'] = i + 'a';
for(int i = 0; m = i, s[i]; ++i) t[i] = decode[s[i] - 'a'];
t[m] = 0;
n = m >> 1;
getNxt(t);
getEx(s + m - n, t); //只要可能是明文的部分
int ans = 0;
for(int i = 0; i < n; ++i)
if(ex[i] == n - i) {ans = m - (n - i); break;}
if(ans) s[ans] = t[ans] = 0;
printf("%s%s\n", s, t);
}
return 0;
}