POJ 2192 Zipper

[b]Zipper[/b]

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14193		Accepted: 4992

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
题目大意：输入三个字符串A,B,C,问C能否有A,B中的字符组成,且字符的顺序必须是原来在字符A,B中的顺序。
解题方法：用动态规划，我用了两种方法，其实思想都是一样的。
方法一：

#include <stdio.h>
#include <string>
#include <cstring>
#include <iostream>
using namespace std;

int dp[205][205];

int main()
{
int N;
scanf("%d", &N);
char str1[205], str2[205], str3[405];
for (int i = 1; i <= N; i++)
{
cin>>str1+1>>str2+1>>str3+1;
int nLen1 = strlen(str1 + 1);
int nLen2 = strlen(str2 + 1);
int nLen3 = strlen(str3 + 1);
memset(dp, 0, sizeof(dp));
int Max = -1;
for (int j = 0; j <= nLen1; j++)
{
for (int k = 0; k <= nLen2; k++)
{
if (j > 0 && str1[j] == str3[j + k])
{
dp[j][k] = max(dp[j][k], dp[j - 1][k] + 1);
Max = max(Max, dp[j][k]);
}
if (k > 0 && str2[k] == str3[j + k])
{
dp[j][k] = max(dp[j][k], dp[j][k - 1] + 1);
Max = max(Max, dp[j][k]);
}
}
}
if (nLen3 == Max)
{
printf("Data set %d: yes\n", i);
}
else
{
printf("Data set %d: no\n", i);
}
}
return 0;
}

方法二：

#include <stdio.h>
#include <string>
#include <cstring>
#include <iostream>
using namespace std;

int dp[205][205];

int main()
{
int N;
scanf("%d", &N);
char str1[205], str2[205], str3[405];
for (int i = 1; i <= N; i++)
{
cin>>str1+1>>str2+1>>str3+1;
int nLen1 = strlen(str1 + 1);
int nLen2 = strlen(str2 + 1);
int nLen3 = strlen(str3 + 1);
memset(dp, 0, sizeof(dp));
int Max = -1;
dp[0][0] = 1;
for (int j = 0; j <= nLen1; j++)
{
for (int k = 0; k <= nLen2; k++)
{
if (j > 0 && str1[j] == str3[j + k] && dp[j - 1][k])
{
dp[j][k] = 1;
}
if (k > 0 && str2[k] == str3[j + k] && dp[j][k - 1])
{
dp[j][k] = 1;
}
}
}
if (dp[nLen1][nLen2])
{
printf("Data set %d: yes\n", i);
}
else
{
printf("Data set %d: no\n", i);
}
}
return 0;
}