hdu-5707-Combine String

                                                          Combine String

                                                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

                                                                     Total Submission(s): 151    Accepted Submission(s): 78


[align=left]Problem Description[/align]
Given three strings a,
b
and c,
your mission is to check whether c
is the combine string of a
and b.

A string c
is said to be the combine string of a
and b
if and only if c
can be broken into two subsequences, when you read them as a string, one equals to
a,
and the other equals to b.

For example, ``adebcf'' is a combine string of ``abc'' and ``def''.

 

[align=left]Input[/align]
Input file contains several test cases (no more than 20). Process to the end of file.

Each test case contains three strings a,
b
and c
(the length of each string is between 1 and 2000).

 

[align=left]Output[/align]
For each test case, print ``Yes'', if
c
is a combine string of a
and b,
otherwise print ``No''.

 

[align=left]Sample Input[/align]

abc
def
adebcf
abc
def
abecdf

 

[align=left]Sample Output[/align]

Yes
No题意：给定a，b，c三个串，问c能否按序分成a和b串，不要求连续。题目链接：Combine String解题思路： 直接模拟，1.统计处a,b,c串各个字符出现的次数，看a,b串的某个字符出现次数的和是否与c串中的相等。 2.从头往后扫描c串，看是否能按序找到a,b串。代码：

//hdu-5707-Combine String

#include
#include
#include
#include

using namespace std;

map mpa,mpb,mpc;
int main()
{
string a,b,c;
int i,lena,lenb,lenc,x,y,flag;
while(cin>>a>>b>>c){
flag=1,x=y=0;
mpa.clear(),mpb.clear(),mpc.clear();
lena=a.length(),lenb=b.length(),lenc=c.length();
for(i=0; i :: iterator it=mpc.begin();
for( ; it!=mpc.end() ;it++){
if(mpa[it->first] + mpb[it->first] != it->second ){  //判断a和b，各个字符出现的次数和c中的是否一样
flag=0;                                          //因为c能且只能分成a和b串
break;
}
}
if(flag && x==lena && y == lenb) cout<<"Yes";   //a和b在c中能按序找到
else                                cout<<"No";
cout<