POJ 1934 Trip LCA及其所有不重复的串

 

Trip

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 1629 Accepted: 372

Description

Alice and Bob want to go on holiday. Each of them has planned a route, which is a list of cities to be visited in a given order. A route may contain a city more than once. 
As they want to travel together, they have to agree on a common route. None wants to change the order of the cities on his or her route or add other cities. Therefore they have no choice but to remove some cities from the route. Of course the common route should be as long as possible. 
There are exactly 26 cities in the region. Therefore they are encoded on the lists as lower case letters from 'a' to 'z'.

Input

The input consists of two lines; the first line is Alice's list, the second line is Bob's list. 
Each list consists of 1 to 80 lower case letters with no spaces inbetween.

Output

The output should contain all routes that meet the conditions described above, but no route should be listed more than once. Each route should be printed on a separate line. There is at least one such non-empty route, but never more than 1000 different ones. Output them in ascending order.

Sample Input

abcabcaaacbacba

Sample Output

ababaabacaabcbaacabaacacaacbaaacbca

Source

CEOI 2003

 

/*以前只是做过求LCA(最大公共子序列)的大小,但是如果要求所有的LCA串,且不允许重复的,这就犯难了.想了n久尝试了n种办法都WA.最后只能去网上看看前辈们的解题报告了,可是这一题解题报告一直没有找到,难怪这题AC的人这么少.搜了n久最后找到了李博CEOI 2003的解题报告，看了之后才恍然大悟.不过这也不是那么容易能想出来的.(1)首先分别计算两个输入串的loc[i][j].str1 的loc[i][j]表示在str1中位于位置i以及它之前的且距离i最近的字母j的位置,则可以得到loc[i][j] = i当 str[i] == j; loc[i - 1][j]当 str[i] != j(2)首先需要对输入串计算一变LCA,即LCA[i][j]表示两个串str1[1...i]与str2[1...j]的LCA的长度 LCA[i][j] = LCA[i - 1][j - 1], 当 str1[i] == str2[j]; max(LCA[i - 1][j], LCA[i][j - 1]), 当 str1[i] != str2[j](2)然后从两个串的末尾递推,假设当前处理到两个串的位置分别为n,m 那么 (1)如果str1[n] == str2[m],则将当前字符str1[n]加入结果字符串res,同时n,m的位置分别向前推进1变为n - 1, m - 1 (2)如果str1[n] != str2[m],则遍历所有的26个字母，找到LCA[loc1[n][k], loc2[m][k]]最大的那些k的位置， 并向前推进，将n, m分别更新为loc1[n][k], loc2[m][k]通过这一题对LCA的认识更加深了一步,继续学习ing*/#include <iostream>#include <string>#include <algorithm>#define MAX_N 10005#define MAX_LEN 100#define maxv(a, b) ((a) >= (b) ? (a) : (b))using namespace std;struct str{ string seq;}strs[MAX_N + 1];int n, maxLen;string str1, str2;int len1, len2;int lcs[MAX_LEN + 1][MAX_LEN + 1];int loc1[MAX_LEN + 1][27];int loc2[MAX_LEN + 1][27];//计算locvoid getLoc(string str, int loc[][27]){ int i, j, len = str.length(); memset(loc, 0, sizeof(loc)); for(i = 1; i <= len; i++) { for(j = 1; j <= 26; j++) { if(str[i - 1] - 'a' + 1 == j) loc[i][j] = i; else { if(loc[i - 1][j] == 0) continue; loc[i][j] = loc[i - 1][j]; } } }}//计算LCSvoid getLCS(){ int i, j; memset(lcs, 0xbf, sizeof(lcs)); for(i = 0; i <= maxv(len1, len2); i++) lcs[i][0] = lcs[0][i] = 0; for(i = 1; i <= len1; i++) { for(j = 1; j <= len2; j++) { if(str1[i - 1] == str2[j - 1]) lcs[i][j] = lcs[i - 1][j - 1] + 1; else { if(lcs[i - 1][j] > lcs[i][j - 1]) lcs[i][j] = lcs[i - 1][j]; else if(lcs[i - 1][j] < lcs[i][j - 1]) lcs[i][j] = lcs[i][j - 1]; else lcs[i][j] = lcs[i][j - 1]; } } }}//逆向处理void solve(int p1, int p2, string res){ //处理到头了则需要将当前的res加入结果字符串数组 if(p1 == -1 || p2 == -1) { strs[n++].seq = res; return; } //当当前位置的两个字符相同时,将这个字符加入结果字符串,并向前推进 if(str1[p1] == str2[p2]) { solve(p1 - 1, p2 - 1, str1[p1] + res); } else { //否则需要找到具有最优(最大)的LCA值的前驱位置 int maxV = INT_MIN, maxP1, maxP2, i; //寻找最大值maxV for(i = 1; i <= 26; i++) { int pp1 = loc1[p1 + 1][i], pp2 = loc2[p2 + 1][i]; //避免在当前位置发生死循环 if(pp1 - 1 == p1 && pp2 - 1 == p2) continue; if(lcs[pp1][pp2] > maxV) { maxV = lcs[pp1][pp2]; maxP1 = pp1; maxP2 = pp2; } } //从找到的那些最大值继续向前推进 for(i = 1; i <= 26; i++) { int pp1 = loc1[p1 + 1][i], pp2 = loc2[p2 + 1][i]; if(pp1 - 1 == p1 && pp2 - 1 == p2) continue; if(lcs[pp1][pp2] == maxV) solve(pp1 - 1, pp2 - 1, res); } }}bool compare(const str &s1, const str &s2){ return s1.seq <= s2.seq;}int main(){ cin>>str1>>str2; maxLen = INT_MIN; n = 0; len1 = str1.length(); len2 = str2.length(); string res = ""; getLoc(str1, loc1); getLoc(str2, loc2); getLCS(); solve(len1 - 1, len2 - 1, res); sort(strs, strs + n, compare); for(int i = 0; i < n; i++) { cout<<strs[i].seq<<endl; res = strs[i].seq; } return 0;}