HDU 4451 Dressing（计数）

Dressing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3807    Accepted Submission(s): 1722

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

题意：衣服裤子鞋子分别有n,m,k件，指定某件衣服跟裤子不能搭配，或者某件裤子跟某双鞋子不能搭配，这个的指定有P条，问你有几种不同的搭配方式。

解题思路：用两个数组 c[maxn] 和 s[maxn] 分别表示第 i 条裤子与多少件衣服不可搭配和多少条裤子与多少双鞋子不可搭配（就是以裤子为中间媒介）。最后只要遍历
一遍所有的裤子，（n-c[i])*(k-s[i])就是每条裤子在满足条件的情况下有多少种搭配方式，最后将所有的搭配求和为ans即可。

<span style="font-size:18px;">#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
const int maxn=1005;
bool vis1[maxn][maxn],vis2[maxn][maxn];
int c[maxn],s[maxn];
int main()
{
    int n,m,k,p,a,b;
    char str1[10],str2[10];
    while(~scanf("%d %d %d",&n,&m,&k) && n+m+k)
    {
        memset(vis1,false,sizeof(vis1));
        memset(vis2,false,sizeof(vis2));
        memset(c,0,sizeof(c));
        memset(s,0,sizeof(s));
        scanf("%d",&p);
        for(int i=0;i<p;i++){
            scanf("%s %d %s %d",&str1,&a,&str2,&b);
            if(str1[0]=='c' && str2[0]=='p')
            {
                if(!vis1[a][b]){
                    vis1[a][b]=true;
                    c[b]++;
                }
            }
            else{
                if(!vis2[a][b]){
                    vis2[a][b]=true;
                    s[a]++;
                }
            }
        }
        int ans=0;
        for(int i=1;i<=m;i++)
            ans+=(n-c[i])*(k-s[i]);
        printf("%d\n",ans);
    }
    return 0;
}
</span>