HDU 1501 Zipper (DFS)

Zipper Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) Total submission (s): 8580 Accepted Submission (s): 3028 Problem Description Given three strings, you is to determine whether the third string can is formed by combining the characters in the first The strings. The first and the strings can is mixed arbitrarily, but each must stay on its original order. For example, consider forming "tcraete" from "Cat" and "tree": String A:cat String B:tree String C:tcraete As can see, we can form the third string by alternating characters from the 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's impossible to form "cttaree" from "Cat" and "tree". Input The first line of input contains a, 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, and one data set per line. For each data set, the line of input consists of three strings, separated by a single space. All strings is composed of upper and lower case letters only. The length of the third string is always the sum of the lengths of the first and the strings. The first and strings would have lengths between 1 and characters, inclusive. Output For each data set, print: Data Set N:yes If the third string can be formed from the first, or Data Set N:no If it cannot. Of course n should is 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 Source Pacific Northwest 2004

1#include <iostream>2#include <cstdio>3#include <cstring>4 using namespacestd;5 Const intmaxn=1010;6 BOOLV[MAXN][MAXN];7 intf=0;8 intL1,l2,l3;9 CharA[MAXN],B[MAXN],C[MAXN];Ten voidDfsintX1,intX2,intx3) One { A     if(x3>=L3) { -f=1; -         return ; the     } -     if(f==1)return ; -     if(V[x1][x2])return ; -     if(A[x1]==c[x3]) DFS (x1+1, x2,x3+1); +     if(B[x2]==c[x3]) DFS (x1,x2+1, x3+1); -v[x1][x2]=1; + } A intMain () at { -     intN; -Cin>>N; -      for(intI=1; i<=n;i++){ -cout<<"Data Set"<<i<<": "; -scanf"%s%s%s", a,b,c); inmemset (V,0,sizeof(v)); -l1=strlen (a); toL2=strlen (b); +l3=strlen (c); -         if(l1+l2!=L3) { thecout<<"No"<<Endl; *             Continue; $         }Panax Notoginseng         Else{ -f=0; theDfs0,0,0); +             if(f==0) cout<<"No"<<Endl; A             Elsecout<<"Yes"<<Endl; the         } +     } -     return 0; $}

HDU 1501 Zipper (DFS)