poj 2250 Compromise(动归）

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#include<iostream>#include<vector>#include<fstream>#include<stdio.h>#include<algorithm>#include<string.h>#include<string>using namespace std;vector<string> p1, p2;vector<string> p[105][105];char sample[35];string word;int numOfMark = 1;void deal();int main(){freopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);p1.push_back("");p2.push_back("");while(scanf("%s",sample)!=EOF){word = sample;if(word == "#"){if(numOfMark &1){++numOfMark;continue;}else{deal();p1.clear();p2.clear();p1.push_back("");p2.push_back("");for(int q = 0;  q<105; ++q){for(int w = 0; w < 105; ++w){p[q][w].clear();}}numOfMark = 1;continue;}}if(numOfMark &1){  //奇数p1.push_back(word);}else{p2.push_back(word);}}return 0;}void deal(){int len1 = p1.size() , len2 = p2.size(), templen1, templen2;for(int i = 1; i < len1; ++i){for(int j = 1; j < len2; ++j){if(p1[i] == p2[j]){p[i][j] = p[i-1][j-1];p[i][j].push_back(p1[i]);}else{templen1 = p[i-1][j].size();templen2 = p[i][j-1].size();if(templen1 > templen2){p[i][j] = p[i-1][j];}else{p[i][j] = p[i][j-1];}}}}int same = p[len1-1][len2-1].size(),limit1= len1-1, limit2= len2-1;for(int i = 0; i < same; ++i){if(i ){printf(" ");}printf("%s",p[limit1][limit2][i].c_str());}printf("\n");printf("\n");return;}

这是一道LCS的变形题，要求两个序列（单词）的最长公共子序列；

首先要知道怎么求LCS(Largest common subsequence）

设有两个序列s1,s2；长度分别为l1,l2；设p[i][j](0 <= i < l1, 0 <= j < l2)是s1的前i个字符与s2的前j个字符的公共子串长度

那么p[i][j] = p[i-1][j-1]+1 当s1[i] == s2[j];

max(p[i][j-1],p[i-1][j]} else

这道题就是把长度换为相同单词的集合

如有语言表述不准确的地方，欢迎指正~谢谢大家