Zipper
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4884 Accepted Submission(s): 1742
Problem Description
Given 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".
Input
The 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, inclusive.
Output
For 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 should be replaced by the data set number. See the sample output below for an example.
Sample Input
3
cat tree tcraete
cat tree catrtee
cat tree cttaree
Sample Output
Data set 1: yes
Data set 2: yes
Data set 3: no
其實還是LCS算法的思想。
opt[i][j] 表示 字符串2的前i個字符 和 字符串1 的前j個字符，可以匹配 字符串3 的最大個數。
例如 對於第一個case(省略第0行和第0列)， 參看代碼:
Data set 1:
2 3 3 3
2 4 5 5
2 4 6 7
狀態方程：
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] ]) );
[cpp]
#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;
else
opt[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;
else
opt[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");
}else
printf("no\n");
}
return 0;
}
#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;
else
opt[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;
else
opt[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");
}else
printf("no\n");
}
return 0;
}
