Compromise
Time Limit: 3000ms Memory Limit: 131072KB
[PDF Link]
In a few months the European Currency Union will become a reality. However, to join the club, the Maastricht criteria must be fulfilled, and this is not a trivial task for the countries (maybe except for Luxembourg). To enforce that Germany will fulfill the criteria, our government has so many wonderful options (raise taxes, sell stocks, revalue the gold reserves,…) that it is really hard to choose what to do.
Therefore the German government requires a program for the following task:
Two politicians each enter their proposal of what to do. The computer then outputs the longest common subsequence of words that occurs in both proposals. As you can see, this is a totally fair compromise (after all, a common sequence of words is something what both people have in mind).
Your country needs this program, so your job is to write it for us.
Input SpecificationThe input file will contain several test cases.
Each test case consists of two texts. Each text is given as a sequence of lower-case words, separated by whitespace, but with no punctuation. Words will be less than 30 characters long. Both texts will contain less than 100 words and will be terminated by a line containing a single ‘#’.
Input is terminated by end of file.
Output SpecificationFor each test case, print the longest common subsequence of words occuring in the two texts. If there is more than one such sequence, any one is acceptable. Separate the words by one blank. After the last word, output a newline character.Sample Inputdie einkommen der landwirtesind fuer die abgeordneten ein buch mit sieben siegelnum dem abzuhelfenmuessen dringend alle subventionsgesetze verbessert werden#die steuern auf vermoegen und einkommensollten nach meinung der abgeordnetennachdruecklich erhoben werdendazu muessen die kontrollbefugnisse der finanzbehoerdendringend verbessert werden#Sample Outputdie einkommen der abgeordneten muessen dringend verbessert werden
题目大意：
给你两段字符串，多行输入，单词之间空格隔开，每段以#号结束，，要求输出这两段的最长的公共子序列。
题目大意：
LCS算法+打印路径，增加path【】【】数组。
代码：
#include<iostream>#include<string>#include<cstring>#include<vector>using namespace std;const int maxn = 105;string s1[maxn],s2[maxn];int m,n,dp[maxn][maxn],path[maxn][maxn];vector <string> answer;void LcsLength(){memset(dp, 0, sizeof(dp));memset(path, 0, sizeof(path));for(int i=1;i<=m;i++){for(int j=1;j<=n;j++){if(s1[i-1]==s2[j-1]) {dp[i][j]=dp[i-1][j-1]+1;path[i-1][j-1]=1;//path[m][n]没有被赋值，只跟新到了path[m-1][n-1],利用#号多读一个，就可以访问到最后一个单词了}else{if(dp[i-1][j]>=dp[i][j-1]) {dp[i][j]=dp[i-1][j];path[i-1][j-1]=2;}else {dp[i][j]=dp[i][j-1];path[i-1][j-1]=3;}}}}}void LCSPrint(int i,int j){if(i<0||j<0) return ;if(path[i][j]==1){LCSPrint(i-1,j-1);answer.push_back(s1[i]);}else if(path[i][j]==2){LCSPrint(i-1,j);}else LCSPrint(i,j-1);}void output(){for(int i=0;i<answer.size();i++){if(i!=0) cout<<" ";cout<<answer[i];}cout<<endl;}int main(){int casen=0,r=0;while(cin>>s1[0]){answer.clear();for(int i=1;;i++){cin>>s1[i];m=i;if(s1[i]=="#") break;}for(int i=0;;i++){cin>>s2[i];n=i;if(s2[i]=="#") break;}LcsLength();LCSPrint(m-1,n-1);//把#号丢掉output();}return 0;}
然后拍一些美得想哭的照片，留给老年的自己。
