Dressing
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2674 Accepted Submission(s): 1179
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.
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.
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
今天上午陪队友一起做的题目之中的一个，我之前没看这题，后来才看的；
思路：先记录好不和谐的衣服和裤子与裤子和鞋子。再枚举衣服和裤子，假设衣服和裤子能够搭配就看鞋子。然后累加能够搭配的就ok了
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
using namespace std;
int a[1010][1010], b[1010][1010], c[1010];
int main()
{
int N, M, K, P;
char str1[10], str2[10];
int t1, t2;
while(scanf("%d %d %d", &N, &M, &K)==3, N || M || K)
{
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
memset(c, 0, sizeof(c));
scanf("%d", &P);
while(P--)
{
scanf("%s %d %s %d", &str1, &t1, &str2, &t2);
if(strcmp(str1,"clothes")==0)
{
a[t1][t2]=1;
}
else
{
b[t1][t2]=1;
c[t1]++;
}
}
int ans=0;
for(int i=1;i<=N;i++)
for(int j=1;j<=M;j++)
if(a[i][j]==0)
ans+=(K-c[j]);
printf("%d
",ans);
}
return 0;
}