(1) meaning of the questions:
N * n have the form of the above different colors balloon, the balloon is the range of colors [1,50]
Every time a person to break the balloons, each can or find a balloon along the line, and the total destruction of all the same color balloons.
A person may operate k times, most asked what color balloons will be left behind.
(2) ideas:
Consider the location where the sum of each color, a color is determined to be completely deleted the minimum number of times num,
Then determine the number of times a certain color to be deleted if more than k, it means that the color will not be deleted.
(3) Code:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<utility>
using namespace std;
const int maxn = 1e4+10;
int vis[maxn],pre[maxn],head[maxn],tot,n,k,Ans[maxn],bj[maxn];
struct Node{
int v,nxt;
}cur[maxn];
vector <pair <int,int> > vc[55];
void Init(){
memset(head,-1,sizeof(head));
tot = 0;
}
void Add(int x,int y){
cur[tot].nxt = head[x];
cur[tot].v = y;
head[x] = tot++;
}
bool dfs(int x){
for(int i=head[x];i!=-1;i=cur[i].nxt){
int y = cur[i].v;
if(vis[y]==0){
vis[y] = 1;
if(pre[y]==-1||dfs(pre[y])){
pre[y] = x;
return true;
}
}
}
return false;
}
int fun(){
int ans = 0;
memset(pre,-1,sizeof(pre));
for(int i=1;i<=n;i++){
memset(vis,0,sizeof(vis));
if(dfs(i)) ans++;
}
return ans;
}
int main(void){
while(~scanf("%d%d",&n,&k)&&(n+k)){
for(int i=1;i<=50;i++){
vc[i].clear();bj[i] = 0;
}
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
int x;scanf("%d",&x);
bj[x] = 1;
vc[x].push_back(make_pair(i,j));
}
}
int tt = 0;
for(int i=1;i<=50;i++)
if(bj[i]==1&&vc[i].size()>0){
Init();
int len = vc[i].size();
for(int j=0;j<len;j++){
pair <int,int> tp = vc[i][j];
Add(tp.first,tp.second);
}
int num = fun();
if(num>k){
Ans[++tt] = i;
}
}
if(tt==0) printf("-1\n");
else{
for(int i=1;i<=tt;i++){
if(i>1) printf(" ");
printf("%d",Ans[i]);
}
printf("\n");
}
}
return 0;
}