10273 - Eat or Not to Eat?

题意：农场有n头奶牛，每头奶牛都有一个产奶周期ti天（ti<=10），每天产奶ai，每天农场都想杀一头奶牛吃，但希望杀产奶最小的一头，如果有多头产奶最小，则那天将不杀。现问最后剩下几头，杀最后一头牛是第几天。

#include<cstdlib>#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include <algorithm>#include<set>#include<queue>#define LL long long#define inf 0x7fffffff#define E 1e-9#define M 100#define N 1005using namespace std;int ma[N][3000];int vis[N];int num[N];int minv[251];int gcd(int a,int b){  if(a<b)    swap(a,b);  int t;  while(b)    {      t=a%b;      a=b;      b=t;    }  return a;}int lcm(int a,int b){  return a/gcd(a,b)*b;}void add(int a[],int k,int m){  int s=m/k-1;  for(int i=0; i<s; i++)    {      memcpy(a+k*(i+1),a,k*sizeof(int));    }}int main(){#ifndef ONLINE_JUDGE  freopen("ex.in","r",stdin);#endif  int ncase,n;  scanf("%d",&ncase);  while(ncase--)    {      memset(vis,0,sizeof(vis));      scanf("%d",&n);      for (int i=0; i<n; i++)        {          scanf("%d",&num[i]);          for (int j=0; j<num[i]; j++)            scanf("%d",&ma[i][j]);        }      int m=1;      for(int i=0; i<n; i++)        m=lcm(m,num[i]);      for(int i=0; i<n; i++)//变为同大小数组        {          add(ma[i],num[i],m);        }      int flag=1;      int time=0,beef=0,ans=0,v,sub;      while(flag)        {          flag=0;          for(int i=0; i<m; i++)            {              ++time;              v=300;              for(int j=0; j<n; j++)                if(!vis[j])                  {                    if(v>ma[j][i])                      {                        v=ma[j][i];                        minv[v]=1;//不用初始化                        sub=j;                      }                    else if(v==ma[j][i])                      minv[v]++;                  }              if(minv[v]==1)                {                  beef++;                  vis[sub]=1;                  ans=time;                  flag=1;                }            }        }      printf("%d %d\n",n-beef,ans);    }  return 0;}