Input
There will be multiple input cases.Each test case begins with two integers n, k. n is the number of rows and columns of the balloons (1 <= n <= 100), and k is the times that ginving to each student(0 < k <= n).Follow a matrix A of n*n, where Aij denote the color of the ballon in the i row, j column.Input ends with n = k = 0.
Output
For each test case, print in ascending order all the colors of which are impossible to be crashed by a student in k times. If there is no choice, print "-1".
Sample Input
1 1
1
2 1
1 1
1 2
2 1
1 2
2 2
5 4
1 2 3 4 5
2 3 4 5 1
3 4 5 1 2
4 5 1 2 3
5 1 2 3 4
3 3
50 50 50
50 50 50
50 50 50
0 0
Sample Output
-1
1
2
1 2 3 4 5
-1
View Code
1 #include<stdio.h> 2 #include<string.h> 3 int n,m; 4 int map[110][110],a[110],b[110]; 5 int count[100],f[100]; 6 int find(int x,int y) 7 { 8 int i; 9 for(i=1;i<=n;i++) 10 { 11 if(map[x][i]==y&&b[i]==0)//颜色要一致，且没有覆盖过 12 { 13 b[i]=1;//表示此次覆盖 14 if(a[i]==0||find(a[i],y)) 15 { 16 a[i]=x;//形成新的交替 17 return 1; 18 } 19 } 20 } 21 return 0; 22 } 23 int met(int x) 24 { 25 int i,sum; 26 sum=0; 27 memset(a,0,sizeof(a));//每一种颜色a都要初始化 28 for(i=1;i<=n;i++) 29 { 30 memset(b,0,sizeof(b)); 31 if(find(i,x)) 32 sum++; 33 } 34 return sum; 35 } 36 int main() 37 { 38 int i,j,k,num; 39 while(scanf("%d%d",&n,&m),n!=0||m!=0) 40 { 41 memset(map,0,sizeof(map));//图标用来记录气球分布 42 memset(count,0,sizeof(count));//记数，记录各颜色的气球出现的次数 43 memset(f,0,sizeof(f));//记录复合题意的气球 44 for(i=1;i<=n;i++) 45 for(j=1;j<=n;j++) 46 { 47 scanf("%d",&map[i][j]); 48 count[map[i][j]]++; 49 } 50 k=0; 51 for(i=1;i<=50;i++) 52 { 53 if(count[i]&&met(i)>m)//如果i气球出现过，并且满足>m的条件 54 f[k++]=i; 55 } 56 if(k==0) 57 printf("-1\n"); 58 else 59 { 60 for(i=0;i<k-1;i++) 61 printf("%d ",f[i]); 62 printf("%d\n",f[k-1]); 63 } 64 } 65 return 0; 66 }