hduoj-1501【dfs】【dp】

题目链接：点击打开链接

Zipper

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9960    Accepted Submission(s): 3571

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

3cat tree tcraetecat tree catrteecat tree cttaree

 

Sample Output

Data set 1: yesData set 2: yesData set 3: no

大意：给你三个字符串，判断第三个字符串是否是前面两个字符串组成的，且前面两个字符串的字符在第三个字符串中的顺序不变。

/*#include<cstdio>#include<algorithm>#include<cstring>using namespace std;bool flag;char a[222],b[222],c[222*2];int len1,len2,len3;void dfs(int x,int y,int len){if(flag)return ;if(len>=len3){flag=1;return ;}if(a[x]==c[len])dfs(x+1,y,len+1);if(b[y]==c[len])dfs(x,y+1,len+1);}int main(){int t,text=0;scanf("%d",&t);while(t--){scanf("%s%s%s",a,b,c);len1=strlen(a);len2=strlen(b);len3=strlen(c);printf("Data set %d: ",++text);if(len1+len2!=len3){puts("no");continue;}flag=0;if(a[len1-1]==c[len3-1]||b[len2-1]==c[len3-1])dfs(0,0,0);if(flag)puts("yes");elseputs("no");}return 0;}*/#include<cstdio>#include<algorithm>#include<cstring>using namespace std;char a[222],b[222],c[222*2];bool dp[222][222];int main(){int t,text=0;scanf("%d",&t);while(t--){scanf("%s%s%s",a+1,b+1,c+1);int len1=strlen(a+1);int len2=strlen(b+1);int len3=strlen(c+1);//printf("%d--%d--%d\n",len1,len2,len3);//printf("%c--%c--%c--\n",a[len1],b[len2],c[len3]);memset(dp,0,sizeof(dp));for(int i=1;i<=len1;i++){if(a[i]==c[i])dp[i][0]=1;}for(int i=1;i<=len2;i++){if(b[i]==c[i])dp[0][i]=1;}for(int i=1;i<=len1;i++){for(int j=1;j<=len2;j++){if(dp[i-1][j]&&(a[i]==c[i+j]))dp[i][j]=1;if(dp[i][j-1]&&(b[j]==c[i+j]))dp[i][j]=1;}}printf("Data set %d: ",++text);if(dp[len1][len2])puts("yes");elseputs("no");} return 0;}