FZU 1350 Very Simple Problem

 

Very Simple Problem

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Time Limit:1s	Memory limit:32M
Accepted Submit:111	Total Submit:317

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During a preparation of programming contest, its jury is usually faced with many difficult tasks. One of them is to select a problem simple enough to most, if not all, contestants to solve. The difficulty here lies in diverse meanings of the term "simple" amongst the jury members. So, the jury uses the following procedure to reach a consensus: each member weights each proposed problem with a positive integer "complexity rating" (not necessarily different for different problems). The jury member calls "simplest" those problems that he gave the minimum complexity rating, and "hardest" those problems that he gave the maximum complexity rating. The ratings received from all jury members are then compared, and a problem is declared as "very simple", if it was called as "simplest" by more than a half of the jury, and was called as "hardest" by nobody. Input The first line of input file contains integers N and P, the number of jury members and the number of problems. The following N lines contain P integers in range from 0 to 1000 each - the complexity ranks. 1 <= N, P <= 100 Output Output must contain an ordered list of problems called as "very simple", separated by spaces. If there are no such problems, output must contain a single integer 0 (zero). Sample Input 4 4 1 1 1 2 5 900 21 40 10 10 9 10 3 4 3 5 Sample Output 3 Original: Northeastern Europe 2002, Far-Eastern Subregion

 

解题：

        找出每行最大最小的问题，分别存在min数组和max数组。再在min数组中找出超过n/2次的数，而且这个数没有在max数组中的，即是要求的数。输出要按顺序。

#include <iostream> #include <string> using namespace std; void sort(int a[],int n) { int i,j; int k; for(i=1;i<n;i++) { k=a[i]; j=i-1; while(j>=0 && a[j]>k) { a[j+1]=a[j]; j--; } a[j+1]=k; } } void toMinMax(int a[],int n,int b[],int c[]) { int i,j=0;int k=0; int min=1000; int max=0; int lengthA=n; for(i=0;i<lengthA;i++) { if(min>a[i]) min=a[i]; if(max<a[i]) max=a[i]; } for(i=0;i<lengthA;i++) { if(min==a[i]) { b[j]=i+1; j++; } if(max==a[i]) { c[k]=i+1; k++; } } } bool isValue(int a[][101],int n,int m,int p) { int length=m; for(int i=0;i<length;i++) for(int j=0;j<p;j++) if(a[i][j]==n) return true; return false; } int howValue(int a[][101],int n,int m,int p) { int length=m; int count=0; for(int i=0;i<length;i++) for(int j=0;j<p;j++) if(a[i][j]==n) count++; return count; } int main() { int N,P,i,j,k=0; int a[101]; int b[101]; int min[101][101]; int max[101][101]; while(cin>>N>>P) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(min,0,sizeof(min)); memset(max,0,sizeof(max)); for(i=0;i<N;i++) { for(j=0;j<P;j++) cin>>a[j]; toMinMax(a,P,min[i],max[i]); } for(i=0;i<N;i++) for(j=0;j<P;j++) { if(howValue(min,min[i][j],N,P)>N/2 && isValue(max,min[i][j],N,P)==false) { int flag=0; for(int m=0;m<k;m++) { if(b[m]==min[i][j]) { flag=1; break; } } if(flag==0) { b[k]=min[i][j]; k++; } } } if(k!=0) { sort(b,k); cout<<b[0]; for(i=1;i<k;i++) cout<<" "<<b[i]; } else cout<<0; cout<<endl; k=0; } return 0; }