Project Euler 81：Path sum: two ways 路径和：两个方向

Path sum: two ways

In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by only moving to the right and down, is indicated in bold red and is equal to 2427.

131	673	234	103	18
201	96	342	965	150
630	803	746	422	111
537	699	497	121	956
805	732	524	37	331

Find the minimal path sum, in matrix.txt (right click and “Save Link/Target As…”), a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right by only moving right and down.

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路径和：两个方向

在如下的5乘5矩阵中，从左上方到右下方始终只向右或向下移动的最小路径和为2427，由标注红色的路径给出。

131	673	234	103	18
201	96	342	965	150
630	803	746	422	111
537	699	497	121	956
805	732	524	37	331

在这个31K的文本文件matrix.txt（右击并选择“目标另存为……”）中包含了一个80乘80的矩阵，求出从该矩阵的左上方到右下方始终只向右和向下移动的最小路径和。

解题

这个题目很简单的

对第0列和第0行的数直接向下加

第0列：data[i][0] = data[i][0] + data[i-1][0]  for i in 1:row - 1

第0行: data[0][i] = data[0][i] + data[0][i-1] for i in 1:col-1

其他情况

for i in 1:row -1

for j in 1:col-1

data[i][j] = data[i][j] + min(data[i-1][j],data[i][j-1])

最后元素data[row-1][col-1]就是最小路径的值。

Python

import time ;
import numpy as np

def run():
    filename = 'E:/java/projecteuler/src/Level3/p081_matrix.txt'
    data = readData(filename)
    Path_Sum(data)
    
def Path_Sum(data):
    row,col = np.shape(data)
    for i in range(1,row):
        data[0][i] = data[0][i]+data[0][i-1]
        data[i][0] = data[i][0] + data[i-1][0]
    for i in range(1,row):
        for j in range(1,col):
            data[i][j] += min(data[i-1][j],data[i][j-1])
    print data[row-1][col-1] 
    
def readData(filename):
    fl = open(filename)
    data =[]
    for row in fl:
        row = row.split(',')
        line = [int(i) for i in row]
        data.append(line)
    return data
if __name__=='__main__':
    t0 = time.time()
    run()
    t1 = time.time()
    print "running time=",(t1-t0),"s"

# 427337
# running time= 0.00799989700317 s
            

参考博客中的读取文件，这个读取文件的思想很好的，自己对于读取文件还不是很熟悉

上个Python程序是按照左上到右下走的

下面java的是按照右下向左上走的

package Level3;

import java.awt.List;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;


public class PE081{
    
    static int[][] grid;
    static void run() throws IOException{
        String filename = "src/Level3/p081_matrix.txt";
        String lineString = "";
        ArrayList<String> listData = new ArrayList<String>();
        BufferedReader data = new BufferedReader(new FileReader(filename));
        while((lineString = data.readLine())!= null){
            listData.add(lineString);
        }
        // 分配大小空间的 定义的grid 没有定义大小
        assignArray(listData.size());
        // 按照行添加到数组grid中 
        for(int index = 0,row_counter=0;index <=listData.size() - 1;++index,row_counter++){
            populateArray(listData.get(index),row_counter);
        }
        System.out.println(Path_min(grid));
        
    }
    public static int Path_min(int[][] data){
        int size = data.length;
        for(int i=size -2;i>=0;--i){
            data[i][size-1] += data[i+1][size-1];
            data[size-1][i] += data[size-1][i+1];
        }
        for( int index = size -2;index >=0;index--){
            for(int innerIndex = size -2;innerIndex >=0;innerIndex--){
                data[index][innerIndex] += Math.min(data[index+1][innerIndex],
                        data[index][innerIndex+1]);
            }
        }
        return data[0][0];
    }
    // 每行的数据添加到数组中
    public static void populateArray(String str,int row){
        int counter = 0;
        String[] data = str.split(",");
        for(int index = 0;index<=data.length -1;++index){
            grid[row][counter++] = Integer.parseInt(data[index]);
        }
    }
    public static void assignArray(int no_of_row){
        grid = new int[no_of_row][no_of_row];
    }
    
    

    public static void main(String[] args) throws IOException{
        long t0 = System.currentTimeMillis();
        run();
        long t1 = System.currentTimeMillis();
        long t = t1 - t0;
        System.out.println("running time="+t/1000+"s"+t%1000+"ms");
//        427337
//        running time=0s38ms
    }
}