hdu 4451(Dressing)

Problem Description

Wangpeng has N clothes, M pants and K shoes so theoretically he can have N×M×K different combinations of dressing.
One day he wears his pants Nike, shoes Adiwang to go to school happily. When he opens the door, his mom asks him to come back and switch the dressing. Mom thinks that pants-shoes pair is disharmonious because Adiwang is much better than Nike. After being asked to switch again and again Wangpeng figure out all the pairs mom thinks disharmonious. They can be only clothes-pants pairs or pants-shoes pairs.
Please calculate the number of different combinations of dressing under mom’s restriction.

 

 

Input

There are multiple test cases.
For each case, the first line contains 3 integers N,M,K(1≤N,M,K≤1000) indicating the number of clothes, pants and shoes.
Second line contains only one integer P(0≤P≤2000000) indicating the number of pairs which mom thinks disharmonious.
Next P lines each line will be one of the two forms“clothes x pants y” or “pants y shoes z”.
The first form indicates pair of x-th clothes and y-th pants is disharmonious(1≤x≤N,1 ≤y≤M), and second form indicates pair of y-th pants and z-th shoes is disharmonious(1≤y≤M,1≤z≤K).
Input ends with “0 0 0”.
It is guaranteed that all the pairs are different.

 

 

Output

For each case, output the answer in one line.

 

 

Sample Input

2 2 2

0

2 2 2

1

clothes 1 pants 1

2 2 2

2

clothes 1 pants 1

pants 1 shoes 1

0 0 0

 

 

Sample Output

8

6

5

 

 

Source

2012 Asia JinHua Regional Contest

 

 

 

Problem analysis:

I learn a lot  from the problem.  When we in the face of a problem,we should observe and analysis the problem first.You can slow but ensure the train of

thought about the problem is right;

For this problem,if you find that the  pants have connections with both clothes and shoes.

Then the problem is not a problem.

#include<iostream>
#include<string>
using namespace std;
#define n 1005
int N,M,K;
int c[n],s[n];
int main()
{
    int i,j;
    char str1[30],str2[30];
    while(scanf("%d%d%d",&N,&M,&K)&&(N+M+K))
    {
        int p;
        for(i=1;i<=M;i++)
        {
            c[i]=N;
            s[i]=K;
        }
        cin>>p;
        getchar();
        int n1,n2;
        for(i=1;i<=p;i++)
        {
            scanf("%s%d%s%d",str1,&n1,str2,&n2);
            if(str1[0]=='c')
                c[n2]--;
            else
                s[n1]--;
        }
        int count=0;
        for(j=1;j<=M;j++)
        {
            count+=c[j]*s[j];
        }
        cout<<count<<endl;

    }
    return 0;
}