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.
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.
For each case, output the answer in one
line.
#include <stdio.h>
#include <string.h>
#define MAXN 1001
int N,M,K;
int map[MAXN][MAXN];
int out[MAXN];
int main(int argc, char *argv[])
{
char ch1[10],ch2[10];
int x,y;
while( scanf("%d %d %d",&N,&M,&K)!=EOF ){
if(N==0 &&M==0 && K==0)break;
memset(map,1,sizeof(map));
for(int i=1; i<=M; i++){
out[i]=K;
}
int P;
scanf("%d",&P);
while(P--){
scanf("%s %d %s %d",ch1,&x,ch2,&y);
if(ch1[0]==‘c‘ && ch2[0]==‘p‘){
map[x][y]=0;
}
if(ch1[0]==‘p‘ && ch2[0]==‘s‘){
if(out[x]>0)
out[x]--;
}
}
__int64 sum=0;
for(int i=1; i<=N; i++){
for(int j=1; j<=M; j++){
if(map[i][j]){
sum+=out[j];
}
}
}
printf("%I64d\n",sum);
}
return 0;
}
