continue out sam ont balloons floating seve sub comm

http://acm.hdu.edu.cn/showproblem.php?pid=1498

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2687 Accepted Submission(s): 1540

Problem Description On Octorber 21st, HDU 50-year-celebration, 50-color balloons floating around the campus, it‘s so nice, isn‘t it? To celebrate this meaningful day, the ACM team of HDU hold some fuuny games. Especially, there will be a game named "crashing color balloons".

There will be a n*n matrix board on the ground, and each grid will have a color balloon in it.And the color of the ballon will be in the range of [1, 50].After the referee shouts "go!",you can begin to crash the balloons.Every time you can only choose one kind of balloon to crash, we define that the two balloons with the same color belong to the same kind.What‘s more, each time you can only choose a single row or column of balloon, and crash the balloons that with the color you had chosen. Of course, a lot of students are waiting to play this game, so we just give every student k times to crash the balloons.

Here comes the problem: which kind of balloon is impossible to be all crashed by a student in k times.

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

Author 8600

Source “2006校園文化活動月”之“校慶杯”大學生程序設計競賽暨杭州電子科技大學第四屆大學生程序設計競賽

Recommend LL | We have carefully selected several similar problems for you: 1528 1281 1151 1533 2063 題意：給你一個n*n的矩陣，k次可以將每行或每列同顏色消去的操作，求一定不能消去的顏色，若沒有輸出-1； n<=100；；；枚舉每次想要消去的顏色，以同種顏色的格子的坐標構圖，得出最大匹配。

 1 #include <algorithm>
 2 #include <cstring>
 3 #include <cstdio>
 4 
 5 const int N(110);
 6 bool map[N][N],vis[N],linked[N];
 7 int n,k,col[N][N],ans[N],match[N];
 8 
 9 bool find(int u)
10 {
11     for(int v=1;v<=n;v++)
12         if(map[u][v]&&!vis[v])
13         {
14             vis[v]=1;
15             if(!match[v]||find(match[v]))
16             {
17                 match[v]=u;
18                 return true;
19             }
20         }
21     return false;
22 }
23 bool link()
24 {
25     int ret=0;
26     for(int i=1;i<=n;i++)
27     {
28         if(find(i)) ret++;
29         memset(vis,0,sizeof(vis));
30     }
31     memset(match,0,sizeof(match));
32     return ret>k;
33 }
34 
35 int main()
36 {
37     for(;scanf("%d%d",&n,&k)&&n&&k;)
38     {
39         int cnt=0;
40         for(int i=1;i<=n;i++)
41           for(int j=1;j<=n;j++)
42             scanf("%d",&col[i][j]);
43         for(int i=1;i<=n;i++)
44             for(int j=1;j<=n;j++)
45               {
46                   if(linked[col[i][j]]) continue;
47                   linked[col[i][j]]=true;
48                   for(int u=1;u<=n;u++)
49                       for(int v=1;v<=n;v++)
50                           if(col[u][v]==col[i][j])
51                               map[u][v]=1;
52                   if(link()) ans[++cnt]=col[i][j];
53                   memset(map,0,sizeof(map));
54             }
55         memset(linked,0,sizeof(linked));
56         if(!cnt) puts("-1");
57         else
58         {
59             std:: sort(ans+1,ans+cnt+1);
60             for(int i=1;i<cnt;i++)
61                 printf("%d ",ans[i]);
62             printf("%d\n",ans[cnt]);
63         }
64         memset(ans,0,sizeof(ans));
65     }
66     return 0;
67 }

HDU——T 1498 50 years, 50 colors