zoukankan      html  css  js  c++  java
  • 【习题 6-8 UVA

    【链接】 我是链接,点我呀:)
    【题意】

    在这里输入题意

    【题解】

    写两个dfs模拟就好。 注意每12个数字输出一个换行。。

    【代码】

    /*
      	1.Shoud it use long long ?
      	2.Have you ever test several sample(at least therr) yourself?
      	3.Can you promise that the solution is right? At least,the main ideal
      	4.use the puts("") or putchar() or printf and such things?
      	5.init the used array or any value?
      	6.use error MAX_VALUE?
      	7.use scanf instead of cin/cout?
    */
    #include <bits/stdc++.h>
    using namespace std;
    
    const int N = 100;
    
    int n;
    char s[N][N];
    vector <long long> v;
    
    int numberblack(int x1,int y1,int x2,int y2){
    	int cnt = 0;
        for (int i = x1;i <= x2;i++)
        	for (int j = y1;j<= y2;j++)
        		if (s[i][j]=='1')
        			cnt++;
        return cnt;
    }	
    
    void dfs(int x1,int y1,int x2,int y2,int now,int xishu){
    	int temp1 = numberblack(x1,y1,x2,y2);
    	if (temp1==0) return;
    	int should = (x2-x1+1)*(y2-y1+1);
    	if (temp1==should){
    	 	v.push_back(now);
    	 	return;
    	}
    	int midx = (x1+x2)/2,midy = (y1+y2)/2;
    	dfs(x1,y1,midx,midy,now + 1*xishu,xishu*5);
    	dfs(x1,midy+1,midx,y2,now + 2*xishu,xishu*5);
    	dfs(midx+1,y1,x2,midy,now + 3*xishu,xishu*5);
    	dfs(midx+1,midy+1,x2,y2,now + 4*xishu,xishu*5);
    }
    
    void fugai(int x1,int y1,int x2,int y2){
     	for (int i = x1;i <= x2;i++)
     		for (int j = y1;j <= y2;j++)
     			s[i][j] = '*';
    }
    
    void dfs2(int x1,int y1,int x2,int y2,int rest){
     	if (rest==0){
    		fugai(x1,y1,x2,y2);
    		return; 	 	
     	}
     	int ope = rest%5;
    
     	int midx = (x1+x2)/2,midy = (y1+y2)/2;
    	switch (ope){
    	 	case 1: dfs2(x1,y1,midx,midy,rest/5);break;
    	 	case 2: dfs2(x1,midy+1,midx,y2,rest/5);break;
    	 	case 3: dfs2(midx+1,y1,x2,midy,rest/5);break;
    	 	case 4: dfs2(midx+1,midy+1,x2,y2,rest/5);break;
    	}
    }
    
    int main(){
    	#ifdef LOCAL_DEFINE
    	    freopen("F:\c++source\rush_in.txt", "r", stdin);
    	    freopen("F:\c++source\rush_out.txt", "w", stdout);
    	#endif
    	ios::sync_with_stdio(0),cin.tie(0);
    	int kase = 0;
    	while (cin >>n && n!=0){
    		if (kase>0) cout << endl;
    		kase++;
    		cout <<"Image "<<kase<<endl;
    		if (n>0){
    			v.clear(); 
    			for (int i = 1;i <= n;i++) cin >> (s[i]+1);
    			dfs(1,1,n,n,0,1); 	
    			sort(v.begin(),v.end());
    			for (int i = 0;i < (int) v.size();i++){
    			 	cout << v[i];
    			 	if ((i+1)%12==0 || i==(int) v.size()-1) cout << endl;else cout <<' ';
    			}
    			cout <<"Total number of black nodes = "<<(int) v.size()<<endl;
    		}else{
    			n = -n;
    		 	for (int i = 1;i <= n;i++)
    		 		for (int j =1 ;j <= n;j++) s[i][j] = '.';
    		 	int x;
    		 	for (cin >>x;x!=-1;cin>>x){
    		 	 	dfs2(1,1,n,n,x);
    		 	}
    
    		 	for (int i = 1;i <= n;i++){
    		 		for (int j =1 ;j <= n;j++) cout <<s[i][j];
    		 		cout << endl;
    			}
    		}
    	}
    	return 0;
    }
    
  • 相关阅读:
    log4j日志基本配置
    MyBatis基本应用
    Java properties配置文件
    Java DAO模式
    Java方式 MySQL数据库连接
    bug 复制代码没有审查,没有完全就该变量名
    cocos3 深入理解tiledmap
    cocos3 深入理解单例模式
    cocos3 CC_BREAK_IF(m_pGameMap==NULL);
    cocos2d3 宏定义屏幕宽高,这样就不用重复获取了
  • 原文地址:https://www.cnblogs.com/AWCXV/p/7894749.html
Copyright © 2011-2022 走看看