Zipper
Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 4884 accepted submission (s): 1742
Problem descriptiongiven three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. the first two strings can be mixed arbitrarily, but each must stay in its original order.
For example, consider forming "tcraete" from "cat" and "Tree ":
String A: Cat
String B: Tree
String C: tcraete
As you can see, we can form the third string by alternating characters from the two strings. As a second example, consider forming "catrtee" from "cat" and "Tree ":
String A: Cat
String B: Tree
String C: catrtee
Finally, notice that it is impossible to form "cttaree" from "cat" and "tree ".
Inputthe first line of input contains a single positive integer from 1 through 1000. it represents the number of data sets to follow. the processing for each data set is identical. the data sets appear on the following lines, one data set per line.
For each data set, the line of input consists of three strings, separated by a single space. all strings are composed of upper and lower case letters only. the length of the third string is always the sum of the lengths of the first two strings. the first two
Strings will have lengths between 1 and 200 characters, intrusive.
Outputfor each data set, print:
Data Set N: Yes
If the third string can be formed from the first two, or
Data Set N: No
If it cannot. Of course n shocould be replaced by the data set number. See the sample output below for an example.
Sample Input
3cat tree tcraetecat tree catrteecat tree cttaree
Sample output
Data set 1: yesData set 2: yesData set 3: no
In fact, it is still the idea of the LCS algorithm.
OPT [I] [J] indicates the first I character of string 2 and the first J character of string 1. It can match the maximum number of character 3.
For example, for the first case (Omitted 0th rows and 0th columns), see the code:
Data Set 1:
2 3 3 3
2 4 5 5
2 4 6 7
State equation:
OPT [I] [J] = max (OPT [I-1] [J] + (str1 [I-1] = str3 [OPT [I-1] [J]), OPT [I] [J-1] + (str2 [J-1] = str3 [OPT [I] [J-1]);
#include <stdio.h>#include <string.h>#include <iostream>using namespace std;int max(int a,int b){return a > b ? a:b;}int main(){//freopen("in.txt","r",stdin);int t;scanf("%d",&t);char str1[210],str2[210],str3[410];int opt[410][410];for(int c=1; c<=t; c++){printf("Data set %d: ",c);scanf("%s %s %s", str1,str2,str3);int len1 = strlen(str1);int len2 = strlen(str2);int k = 0;opt[0][0] = 0;for(int i=1; i<=len2; i++){if(str2[i-1] == str3[i-1])opt[0][i] = opt[0][i-1] + 1;elseopt[0][i] = opt[0][i-1];}for( i=1; i<=len1; i++){if(str1[i-1] == str3[i-1])opt[i][0] = opt[i-1][0] + 1;elseopt[i][0] = opt[i-1][0];}for( i=1; i<=len1; i++){for(int j=1; j<=len2; j++){opt[i][j] = max(opt[i-1][j] + (str1[ i-1 ] == str3[ opt[i-1][j] ]), opt[i][j-1] + (str2[ j-1 ] == str3[ opt[i][j-1] ]) );//cout << opt[i][j] << " ";}}if(opt[len1][len2] == len1 + len2){printf("yes\n");}elseprintf("no\n");}return 0;}