华为给出的测试账号里面的题

#include<iostream>

#include<algorithm>

#include<map>

#include<string>

usingnamespace
std;

typedef pair<string,int>MyPair;

typedefmap<string,int>MyMap;

map<string,int>::iterator
iter;

int main()

{

int count=
0;

const string add=
"addCandidate";

const string vote=
"vote";

const string get=
"getVoteResult";

string cmdf;

MyMap mymap;

while(cmdf
!= get)

{

string cmdn;

string str;

cout<<
"请输入命令"<< endl;

cin>> cmdf;

if(cmdf
!= get)

cin>> cmdn;

if(cmdf
== add)

{

str.clear();

for(int
i=
0; i< cmdn.size();
i++)

{

str.push_back(cmdn[i]);

}

mymap.insert(make_pair(str,0));

}

elseif
(cmdf== vote)

{

str.clear();

for(int
i=
0; i< cmdn.size();
i++)

str.push_back(cmdn[i]);

iter= mymap.find(str);

if(iter
!= mymap.end())

iter->second++;

else

count++;

}

}

iter= mymap.begin();

while(iter
!= mymap.end())

{

cout << (*iter).first << " " << (*iter).second << endl;

iter++;

}

cout<<
"输错次数："<< count
<< endl;

return0;

}

这个题目开始想用回溯法做，用的时间较多，后来发现这种暴力破解法貌似更好，题目没有时间，内存的限制~

#include<iostream>

usingnamespace
std;

void old_Home(char**m,int
x,int y,int&
a,int& b)//获取B的坐标

{

for(int
i=
0; i< x; i++)

{

for(int
j=
0; j< y; j++)

{

if(m[i][j]==
'B')

{

a= i;

b= j;

   }

}

}

}

bool find_Home(char**m,int
x,int y,int
a,int b)

{

if((a
>= 0)&&
(a< x)&&
(b>=
0)&&
(b< y))

{

if(m[a][b]==
'H')

returntrue;

if(m[a][b]==
'#')

returnfalse;

if((m[a][b]==
'-')||
(m[a][b]==
'B'))

{

m[a][b]=
'#';

return(find_Home(m,
x, y, a
+ 1, b)||

find_Home(m,
x, y, a-
1, b)||

find_Home(m,
x, y, a, b+
1)||

find_Home(m,
x, y, a, b-
1));

}

}

else

returnfalse;

}

voidInit(char*
p,int x,int
y)//输入矩阵

{

for(int
i=
0; i< x
; i++)

{

cin>> p;

p+= y;

}

}

int _tmain(int
argc, _TCHAR* argv[])

{

int x,
y,a
= 0,b=
0;

cin>> x;

cin>> y;

char** m=
newchar*[x];

char* p=
newchar[x*y+1];

for(int
i=
0; i< x; i++)

m[i]=
p + i*y;

Init(p,
x, y);

old_Home(m,
x, y, a, b);

if(find_Home(m,
x, y, a, b))

cout<<
"Y"<< endl;

else

cout<<
"N"<< endl;

delete[] p;

return0;

}