POJ 2192 判断两个字符串是否能按顺序组合成第三个字符串 ZIPPER

 

经典动态规划

现在对动态规划的子结构划分和子结构的理解，转移方程还没有明确的认识

Zipper

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9409 Accepted: 3218

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

/////opt[i][j]:表示str1[i],str2[j]能否组合成st3[i+j]用0，1表示；//状态转移方程：若opt[i-1][j],opt[i][j-1]都为0，则opt[i][j]为0；//若opt[i-1][j] ==1;且str3[i+j] == str1[i],则opt[i][j] = 1;//若opt[i][j-1] == 1;且str3[i+j] == str2[j],则opt[i][j] = 1;//初始化，opt[0][0] = 1;//opt[i][0]表示st1[]的前i个字符是否都与str3[i]相符，相符则为1，否则为0；//opt[0][i]表示st2[]的前i个字符是否都与str3[i]相符，相符则为1，否则为0；#include<stdio.h>#define MAX 201int opt[MAX][MAX];int main(){    int n;    int cnt=0;    int flag;    //freopen("input","r",stdin);    char str1[MAX],str2[MAX],str3[MAX*2];    int len_str1,len_str2,len_str3;    int i,j;    scanf("%d",&n);    while(n--)    {        cnt++;        scanf("%s %s %s",str1,str2,str3);        len_str1=strlen(str1);        len_str2=strlen(str2);        len_str3=strlen(str3);        if(len_str1+len_str2!=len_str3)        {            printf("Data set %d: no/n");            continue;        }        memset(opt,0,sizeof(opt));        opt[0][0]=1;        for(i=1;i<=len_str1;i++)        {            if(opt[i-1][0]==1&&str1[i-1]==str3[i-1]) opt[i][0]=1;            else  break;        }        for(j=1;j<=len_str2;j++)        {            if(opt[0][j-1]==1&&str2[j-1]==str3[j-1]) opt[0][j]=1;            else  break;        }        for(i=1;i<=len_str1;i++)        {            for(j=1;j<=len_str2;j++)            {                if((opt[i][j-1]==1&&str2[j-1]==str3[i+j-1])||(opt[i-1][j]==1&&str1[i-1]==str3[i+j-1]))                    opt[i][j]=1;            }        }        if(opt[len_str1][len_str2])  printf("Data set %d: yes/n",cnt);        else printf("Data set %d: no/n",cnt);    }    return 0;}