CodeForces Round #145(234D) - Cinema

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算出每个演出的可能最少喜欢演员人数MIN,和可能最多喜欢演员人数MAX...对于一个演出来说若其MIN不小于其他所有演出的MAX..则其为surely be the favourite...若其MAX不大于其他所有演出的MIN,则其为won't surely be the favourite...都不满足..只能是can either be favourite, or not favourite

Program:

#include<iostream>#include<stdio.h>#include<string.h>#include<cmath>#include<algorithm>#include<stack>#include<queue>#include<set>#define ll long longusing namespace std;struct node{    int MAX,MIN;}c[105];int m,n,num,k,x,a[105];bool f[105];char s[25];int main(){      freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);    int i,j,n1,n2,nn,h;    while (~scanf("%d%d",&m,&k))    {          memset(f,false,sizeof(f));          for (i=1;i<=k;i++)           {                 scanf("%d",&j);                 f[j]=true;          }          scanf("%d",&n);          for (i=1;i<=n;i++)          {                 scanf("%s%d",s,&num);                 n1=k,n2=m-k,h=nn=0;                 for (j=1;j<=num;j++)                  {                        scanf("%d",&x);                         if (!x) nn++;                           else                              if (f[x]) h++,n1--;                                else n2--;                 }                 if (n1>=nn) c[i].MAX=h+nn;                        else c[i].MAX=h+n1;                 if (n2>=nn) c[i].MIN=h;                        else c[i].MIN=h+nn-n2;          }           for (i=1;i<=n;i++)          {                  for (j=1;j<=n;j++)                 if (j!=i && c[j].MAX>c[i].MIN) break;                 if (j>n) printf("0\n");                     else                     {                           for (j=1;j<=n;j++)                           if (j!=i && c[j].MIN>c[i].MAX) break;                            if (j<=n) printf("1\n");                                else printf("2\n");                     }          }    }    return 0;}