[CodeVS]2495:水叮当的舞步

Posted 2020-08-14

传送门

A*搜索练习，以还剩余多少种颜色为估价，写的时候把2打成了1查了一个多小时。

 

//codevs 2495
//by Cydiater
//2016.10.11
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <map>
#include <ctime>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <algorithm>
using namespace std;
#define ll long long
#define up(i,j,n)		for(int i=j;i<=n;i++)
#define down(i,j,n)		for(int i=j;i>=n;i--)
#define pii pair<int,int>
#define mp make_pair
const int MAXN=9;
const int oo=0x3f3f3f3f;
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
inline int read(){
	char ch=getchar();int x=0,f=1;
	while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
	while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
	return x*f;
}
int N,color[MAXN][MAXN],rec[MAXN][MAXN],tmp[MAXN][MAXN],cnt[10];
bool OK,vis[MAXN][MAXN];
namespace solution{
	void init(){
		scanf("%d\n",&N);if(N==0)exit(0);
		memset(rec,0,sizeof(rec));
		up(i,1,N)up(j,1,N)color[i][j]=read();
	}
	void dfs(int x,int y,int col){
		rec[x][y]=1;
		up(i,0,3){
			int tx=x+dx[i],ty=y+dy[i];
			if(tx>N||tx<1||ty>N||ty<1||rec[tx][ty]==1)continue;
			if(color[tx][ty]==col)	dfs(tx,ty,col);
			else 					rec[tx][ty]=2;
		}
	}
	int get(){
		int ans=0;
		memset(cnt,0,sizeof(cnt));
		up(i,1,N)up(j,1,N)if(rec[i][j]!=1&&!cnt[color[i][j]]){
				ans++;cnt[color[i][j]]=1;
		}
		return ans;
	}
	bool fill(int col){
		bool flag=0;
		up(i,1,N)up(j,1,N)if(rec[i][j]==2&&color[i][j]==col){
			flag=1;
			dfs(i,j,color[i][j]);
		}
		return flag;
	}
	void search(int dep,int depth,int now){
		if(OK)return;
		int v=get();
		if(v==0){
			OK=1;
			return;
		}
		if(dep+v>depth)return;
		int ttt[MAXN][MAXN];
		up(i,0,5){
			memcpy(ttt,rec,sizeof(rec));
			if(fill(i))search(dep+1,depth,i);
			memcpy(rec,ttt,sizeof(ttt));
		}
	}
	void slove(){
		dfs(1,1,color[1][1]);
		OK=0;
		up(dep,0,20){
			search(0,dep,color[1][1]);
			if(OK){
				printf("%d\n",dep);
				return;
			}
		}
	}
}
int main(){
	using namespace solution;
	while(1){
		init();
		slove();
	}
	return 0;
}