1110 - An Easy LCS
PDF (English) Statistics Forum
Time Limit: 2 second(s) Memory Limit: 32 MB
LCS means ‘Longest Common Subsequence‘ that means two non-empty strings are given; you have to find the Longest Common Subsequence between them. Since there can be many solutions, you have to print the one which is the lexicographically smallest. Lexicographical order means dictionary order. For example, ‘abc‘ comes before ‘abd‘ but ‘aaz‘ comes before ‘abc‘.
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case starts with a blank line. The next two lines will contain two strings of length 1 to 100. The strings contain lowercase English characters only.
Output
For each case, print the case number and the lexicographically smallest LCS. If the LCS length is 0 then just print ‘:(‘.
Sample Input
Output for Sample Input
3
ab
ba
zxcvbn
hjgasbznxbzmx
you
kjhs
Case 1: a
Case 2: zxb
Case 3: :(
解题思路: 用LCS的模板 加上记录路径就可以了。
但在做这题的过程中提交一直RE， 本地运行没问题。 仔细找之后才发现自己是从i=0，j=0开始遍历的，出现dp[i-1][j-1]的情况就RE，但是dev 不爆:(
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N = 110;
int dp[N][N];
char str1[N],str2[N],ans[N][N][N];
int main(){
int T;
scanf("%d",&T);
for(int t=1;t<=T;t++){
memset(dp,0,sizeof(dp));
memset(ans,0,sizeof(ans));
scanf("%s %s",str1+1,str2+1);
int len1 = strlen(str1+1),len2 = strlen(str2+1);
for(int i=1;i<=len1;i++){
for(int j=1;j<=len2;j++){
if(str1[i]==str2[j]){
dp[i][j] = dp[i-1][j-1] + 1;
strcpy(ans[i][j],ans[i-1][j-1]);
ans[i][j][dp[i-1][j-1]] = str1[i];
ans[i][j][dp[i][j]] = ‘\0‘;
}else{
dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
if(dp[i-1][j]<dp[i][j-1]){
strcpy(ans[i][j],ans[i][j-1]);
}else if(dp[i-1][j]>dp[i][j-1]){
strcpy(ans[i][j],ans[i-1][j]);
}else{
if(strcmp(ans[i][j-1],ans[i-1][j])>0){
strcpy(ans[i][j],ans[i-1][j]);
}else{
strcpy(ans[i][j],ans[i][j-1]);
}
}
}
}
}
if(strlen(ans[len1][len2])==0)printf("Case %d: :(\n",t);// strcpy(ans[len1-1][len2-1],":(\0");
else printf("Case %d: %s\n",t,ans[len1][len2]);
}
}