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  • 【Codeforces 429B】Working out

    【链接】 我是链接,点我呀:)
    【题意】

    两个人,一个人在左上角,一个人在左下角。 左上角要到右下角去 左下角要到右上角去 只能走到相邻的格子(不能往回走,即一个往右下,一个往右上走) 要求这两个人必须要在这个n*m的格子中选择一个格子作为休息点. (两条路径只能有一个一样的点) 问最多能拿到多少分数(每个格子上有一个分数值)

    【题解】

    ![](https://img2018.cnblogs.com/blog/1251265/201902/1251265-20190214171159886-1414539422.png) 会发现只有②和③不会重复走 则设f[p][i][j]分别为从左上、左下、右上、右下出发,到达点(i,j)的最大分数值 枚举休息点在什么地方 根据上图取答案的最大值就好

    【代码】

    import java.io.*;
    import java.util.*;
    
    public class Main {
    	
    	
    	static InputReader in;
    	static PrintWriter out;
    		
    	public static void main(String[] args) throws IOException{
    		//InputStream ins = new FileInputStream("E:\rush.txt");
    		InputStream ins = System.in;
    		in = new InputReader(ins);
    		out = new PrintWriter(System.out);
    		//code start from here
    		new Task().solve(in, out);
    		out.close();
    	}
    	
    	static int N = (int)1e3;
    	static class Task{
    		public void solve(InputReader in,PrintWriter out) {
    			int n,m;
    			int [][]a = new int[N+10][N+10];
    			int [][][]f = new int[4][N+10][N+10];
    			n = in.nextInt();m = in.nextInt();
    			for (int i = 1;i <= n;i++)
    				for (int j = 1;j <= m;j++)
    					a[i][j] = in.nextInt();
    			
    			//f[0]
    			for (int i = 1;i <= n;i++)
    				for (int j = 1;j <= m;j++)
    					f[0][i][j] = Math.max(f[0][i-1][j], f[0][i][j-1])+a[i][j];
    			
    			//f[1]
    			for (int i = 1;i <= n;i++)
    				for (int j = m;j >= 1;j--)
    					f[1][i][j] = Math.max(f[1][i-1][j], f[1][i][j+1])+a[i][j];
    			
    			//f[2]
    			for (int i = n;i >= 1;i--)
    				for (int j = 1;j <= m;j++)
    					f[2][i][j] = Math.max(f[2][i][j-1], f[2][i+1][j])+a[i][j];
    			
    			//f[3]
    			//f[0]
    			for (int i = n;i >= 1;i--)
    				for (int j = m;j >= 1;j--)
    					f[3][i][j] = Math.max(f[3][i+1][j], f[3][i][j+1])+a[i][j];
    			
    			int ans = 0;
    			for (int i = 1;i <= n;i++)
    				for (int j = 1;j <= m;j++) {
    					//(i,j)
    					if (i>=2 && i<=n-1 && j>=2 && j <= m-1) {
    						//plan 2
    						int temp0,temp1,temp2,temp3;
    						temp0 = f[0][i-1][j];
    						temp1 = f[1][i][j+1];
    						temp2 = f[2][i][j-1];
    						temp3 = f[3][i+1][j];
    						ans = Math.max(ans, temp0+temp1+temp2+temp3);
    						
    						//plan 3
    						temp0 = f[0][i][j-1];
    						temp1 = f[1][i-1][j];
    						temp2 = f[2][i+1][j];
    						temp3 = f[3][i][j+1];
    						ans = Math.max(ans, temp0+temp1+temp2+temp3);
    					}
    				}
    			out.println(ans);
    		}
    	}
    
    	
    
    	static class InputReader{
    		public BufferedReader br;
    		public StringTokenizer tokenizer;
    		
    		public InputReader(InputStream ins) {
    			br = new BufferedReader(new InputStreamReader(ins));
    			tokenizer = null;
    		}
    		
    		public String next(){
    			while (tokenizer==null || !tokenizer.hasMoreTokens()) {
    				try {
    				tokenizer = new StringTokenizer(br.readLine());
    				}catch(IOException e) {
    					throw new RuntimeException(e);
    				}
    			}
    			return tokenizer.nextToken();
    		}
    		
    		public int nextInt() {
    			return Integer.parseInt(next());
    		}
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/10375697.html
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