先上题目：

Time limit: 1.0 second
Memory limit: 64 MB

　　It's well-known that the most tenacious of life species on the Earth are cockroaches. They live everywhere if only there in food. And as far as they are unpretentious in food you can find them absolutely everywhere.

　　A little Lyosha studies at school on a Space station. During one of the school competitions his class has reached the final. A task of the final contest is to exterminate all the cockroaches in the cargo module within minimal time.

　　Within the long history of the competitions a unified tactics was worked out. The tactics is as follows: a poison gas is let in one of the module compartments and after that the baffle that separates the compartment from one of the adjacent ones is opened.  Cockroaches can't stand the smell of the gas and run to the other compartment. When there's no cockroaches in the treated compartment the baffle is closed. Afterwards analogously the next compartment is treated, and so on. The goal is to move all the cockroaches to the floodgate of the cargo module. Then the outward door is opened and all the cockroaches are engulfed by an open Space.

　　Lyosha is responsible for programming the control board of the baffles in his team. The baffles are opened slowly, so it's very important to make do with minimal number of baffle openings in order to win in the contest. Your task is to help Lyosha to compute this number.

Input

　　The first line contains a name of the floodgate compartment. Each of the next lines contains description of one of the baffles — the names of two compartments separated with a dash (-). The last line contains the only symbol "#". There are cockroaches in all the compartments of the module at first. It's possible to get to the floodgate from every compartment of the module passing several baffles. The total number of compartments doesn't exceed 30. The name of a compartment consists of no more than 20 Latin letters and digits. The large and the small letters should be distinguished.

Output

　　Your program is to output the only number — the minimal amount of baffles that should be opened (and then closed) in order to move all the cockroaches to the floodgate.

Sample

input

output

Gateway
Machinery-Gateway
Machinery-Control
Control-Central
Control-Engine
Central-Engine
Storage-Gateway
Storage-Waste
Central-Waste
#

6

　　题意：给你一个终点的名字，给你一个连通图上不同点之间的关系，点上都有蟑螂。现在每一次可以选着一条边，将一个点上的蟑螂都赶到边的另一端，然后这条边会消失，问至少需要多少步操作才能将所有的蟑螂赶到终点。

　　其实这就是在问一个联通图上面保留多少条边可以形成一棵树。树的结构是有n个点就有n-1条边。所以只需要记录每一个出现过的端点的名字各一次，树木减一就得出正确答案了。这里可以使用STL的set容器比较简单，当然用map也可以，不过set会快一点，也可以写一棵字典树，不过那样就很费时间了。

上代码：

#include
#include
#include
#include

using namespace std;

typedef struct{
char k[];
}st;

set s;

int main()
{
st a,b,c;
string x,y;
int tot;
//freopen("data.txt","r",stdin);
scanf("%s",a.k);
x=a.k;
getchar();
tot=;
s.clear();
s.insert(x);
tot++;
while(scanf("%s",a.k),strcmp(a.k,"#")){
getchar();
int l=strlen(a.k);
int j=;
for(int i=;i<l;i++){
if(a.k[i]=='-'){
b.k[j]='\0';
strcpy(c.k,a.k+i+);
break;
}
b.k[j]=a.k[i];
j++;
}
x=b.k;
y=c.k;
if(s.count(x)<=){
s.insert(x);
tot++;
}
if(s.count(y)<=){
s.insert(y);
tot++;
}
}
printf("%d\n",tot-);
return ;
}

1213