50 years, 50 colors

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1695    Accepted Submission(s): 931

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校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

Recommend

看了好几遍愣是没看懂题目什么意思....

看了别人的题解才懂这道题要我们干啥。

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cstring>
 4 const int maxn = 110;
 5 int ma[maxn][maxn];
 6 int line[maxn],res[maxn];
 7 bool vis[maxn],bj[maxn];
 8 int k,n,m,t;
 9 template <class T>
10 void init(T a){
11     memset(a,0,sizeof(a));
12 }
13 int match(int st,int u)
14 {
15     for(int v = 1;v<=n;v++)
16     {
17         if(ma[u][v]== st && !vis[v])
18         {
19             vis[v] = 1;
20             if(line[v]==-1 || match(st,line[v]))
21             {
22                 line[v] = u;
23                 return 1;
24             }
25         }
26     }
27     return 0;
28 }
29 int cmp(const void *a,const void *b){
30     return *(int *)a - *(int *)b;
31 }
32 int K_M(int st)
33 {
34     memset(line,-1,sizeof(line));
35     int ans = 0;
36     for(int i = 1;i<=n;i++){
37         init(vis);
38         ans += match(st,i);
39     }
40     return ans;
41 }
42 int main()
43 {
44     int t;
45     while(~scanf("%d%d",&n,&k))
46     {
47         if(n==0 && k== 0) break;
48         init(ma); init(res); init(bj);
49         for(int i = 1;i<=n;i++)
50         {
51             for(int j = 1;j<=n;j++)
52             {
53                 scanf("%d",&t);
54                 ma[i][j] = t;
55                 if(!bj[t])
56                     bj[t] = 1;
57             }
58         }
59         int cnt,num = 0;
60         for(int i = 1;i<=55;i++)
61         {
62             if(bj[i]){
63                  cnt = K_M(i);
64                  (cnt>k)?(res[num++] = i):(1);
65             }
66         }
67         if(num==0) puts("-1");
68         else{
69             qsort(res,num,sizeof(res[0]),cmp);
70             for(int j = 0;j<num-1;j++)
71                 printf("%d ",res[j]);
72             printf("%d\n",res[num-1]);
73         }
74     }
75     return 0;
76 }

hdu----149850 years, 50 colors(最小覆盖点)