usaco[2018DEC]

sliver

Each of his M cows has two favorite pastures. Some of his cows have a dietary restriction that they should only eat one type of grass consistently --- Farmer John therefore wants to make sure the same type of grass is planted in the two favorite fields of any such cow. Other cows have a very different dietary restriction, requiring them to eat different types of grass. For those cows, Farmer John of course wants to make sure their two favorite fields contain different grass types.  

T2

堆

2.23

T3

/*
ID:Li Zhou
LANG: C++
TASK: test
*/
#include<bits/stdc++.h>
using namespace std;
const int dx[5]={0,0,0,1,-1};
const int dy[5]={0,1,-1,0,0};
int n,k,cnt,az;
bool vis[200][11];
char a[200][11];
int b[200][11];
int bfs(int x,int y,int cal){
    if(vis[x][y])return 0;
    int ans=0;
    if(x<=0||x>n||b[x][y]!=cal)return 0;
    if(b[x][y]==0)return 0;
    if(y<=0||y>10)return 0;
    vis[x][y]=1;
    for(int i=1;i<=4;i++){
    int azx=x+dx[i];
    int azy=y+dy[i];
    ans+=bfs(azx,azy,cal);
    }
    return ans+1;
}
void del(int x,int y,int cal){
    if(b[x][y]==cal&&x>=0&&x<=n&&y>=0&&y<=10){
    b[x][y]=0;
    for(int i=1;i<=4;i++){
    int azx=x+dx[i];
    int azy=y+dy[i];
    del(azx,azy,cal);
    }
    }
}
void work(){
    az=1;
    while(az){
    az=0;
    memset(vis,0,sizeof(vis));
    for(int h=1;h<=n;h++){
      if(az)break;
      for(int j=1;j<=10;j++){
          if(az)break;
          bool t1,t2;
          t2=vis[h][j];
          t1=b[h][j]==0;
      if(t1)continue;
      if(t2)continue;
      cnt=bfs(h,j,b[h][j]);
      if(cnt>=k){
          del(h,j,b[h][j]);
          az=1;
      }}}}
        for(int i=1;i<=10;i++){
          for(int j=2;j<=n;j++)
              if(b[j][i]==0){
                for(int k=j-1;k>=1;k--)
                  b[k+1][i]=b[k][i];
                  b[1][i]=0;
          }
          }
}
int main(){
    freopen("mooyomooyo.in","r",stdin);
    freopen("mooyomooyo.out","w",stdout);
    cin>>n>>k;
    for(int i=1;i<=n;i++)
     for(int j=1;j<=10;j++)
     {cin>>a[i][j];b[i][j]=a[i][j]-48;}
    for(int i=1;i<=1000;i++){
    work();
    }
    //if(!az)break;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=10;j++)cout<<b[i][j];
        if(i!=n)cout<<endl;    
    }
    return 0;
}