hdu 4451 Dressing

Dressing

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2659    Accepted Submission(s): 1169

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 2 202 2 21clothes 1 pants 12 2 22clothes 1 pants 1pants 1 shoes 10 0 0

 

Sample Output

865

 

主要是 会在pants限制时 会多计算 所以之后要加回来

还要注意 同一pants重复出现的情况！！！

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <string.h>#include <string>#define eps 1e-8#define op operator#define MOD  10009#define MAXN  100100#define INF 0x7fffffff#define FOR(i,a,b)  for(int i=a;i<=b;i++)#define FOV(i,a,b)  for(int i=a;i>=b;i--)#define REP(i,a,b)  for(int i=a;i<b;i++)#define REV(i,a,b)  for(int i=a-1;i>=b;i--)#define MEM(a,x)    memset(a,x,sizeof a)#define ll __int64using namespace std;int pant[1010];int c[2000000],pc[2000000],ps[2000000],s[2000000];int main(){//freopen("ceshi.txt","r",stdin);    int n,m,k;    while(scanf("%d%d%d",&n,&m,&k)!=EOF)    {        if(n==0&&m==0&&k==0) break;        int p;        scanf("%d",&p);        if(p==0)        {            printf("%d\n",n*m*k);            continue;        }        MEM(pant,0);        string ty1,ty2;        int cn=0,pnc=0,pns=0,sn=0;        for(int i=0;i<p;i++)        {            int x,y;            cin>>ty1>>x>>ty2>>y;            if(ty1[0]=='c')            {                c[cn++]=x; pc[pnc++]=y;                pant[y]++;            }            else            {                ps[pns++]=x; s[sn++]=y;            }        }//        cout<<"pns  "<<pns<<endl;        int sum=n*m*k;        sum-=cn*k;        sum-=sn*n;        for(int i=0;i<pns;i++)        {            if(pant[ps[i]])                sum+=pant[ps[i]];        }        printf("%d\n",sum);    }    return 0;}