/*
 * The MIT License (MIT)
 *
 * Copyright © 2022 xrv <xrg@live.com>
 *
 * Permission is hereby granted, free of charge, to any person obtaining a copy
 * of this software and associated documentation files (the "Software"), to deal
 * in the Software without restriction, including without limitation the rights
 * to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
 * copies of the Software, and to permit persons to whom the Software is
 * furnished to do so, subject to the following conditions:
 *
 * The above copyright notice and this permission notice shall be included in
 * all copies or substantial portions of the Software.
 *
 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
 * IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
 * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
 * AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
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 * OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
 * THE SOFTWARE.
 */

package io.github.ehlxr.algorithm.dp;

/**
 * 假设我们有一个 n 乘以 n 的矩阵 w [n][n]。矩阵存储的都是正整数。棋子起始位置在左上角，终止位置在右下角。
 * 我们将棋子从左上角移动到右下角。每次只能向右或者向下移动一位。从左上角到右下角，会有很多不同的路径可以走。
 * 我们把每条路径经过的数字加起来看作路径的长度。那从左上角移动到右下角的最短路径长度是多少呢？
 *
 * @author ehlxr
 * @since 2022-03-05 16:46.
 */
public class MinDist {

    private final int[][] matrix = {{1, 3, 5, 9}, {2, 1, 3, 4}, {5, 2, 6, 7}, {6, 8, 4, 3}};
    private final int[][] mem = new int[4][4];
    private int minDist = Integer.MAX_VALUE; // 全局变量或者成员变量

    /**
     * 动态规划：状态转移表法
     */
    public int minDistDP(int[][] matrix, int n) {
        int[][] states = new int[n][n];
        int sum = 0;
        for (int j = 0; j < n; ++j) { // 初始化states的第一行数据
            sum += matrix[0][j];
            states[0][j] = sum;
        }
        sum = 0;
        for (int i = 0; i < n; ++i) { // 初始化states的第一列数据
            sum += matrix[i][0];
            states[i][0] = sum;
        }
        for (int i = 1; i < n; ++i) {
            for (int j = 1; j < n; ++j) {
                states[i][j] = matrix[i][j] + Math.min(states[i][j - 1], states[i - 1][j]);
            }
        }
        return states[n - 1][n - 1];
    }

    /**
     * 动态规划：状态转移方程法
     */
    public int minDist(int i, int j) { // 调用minDist(n-1, n-1);
        if (i == 0 && j == 0) {
            return matrix[0][0];
        }
        // 备忘录，如果计算过，即有数字存在，则直接返回
        if (mem[i][j] > 0) {
            return mem[i][j];
        }
        int minLeft = Integer.MAX_VALUE;
        if (j - 1 >= 0) {
            minLeft = minDist(i, j - 1);
        }
        int minUp = Integer.MAX_VALUE;
        if (i - 1 >= 0) {
            minUp = minDist(i - 1, j);
        }

        int currMinDist = matrix[i][j] + Math.min(minLeft, minUp);
        mem[i][j] = currMinDist;
        return currMinDist;
    }

    /**
     * 回溯方式
     */
    public void minDistBacktracing(int i, int j, int dist, int[][] w, int n) {
        // 到达了n-1, n-1这个位置了，这里看着有点奇怪哈，你自己举个例子看下
        if (i == n && j == n) {
            if (dist < minDist) {
                minDist = dist;
            }
            return;
        }
        if (i < n) { // 往下走，更新i=i+1, j=j
            minDistBacktracing(i + 1, j, dist + w[i][j], w, n);
        }
        if (j < n) { // 往右走，更新i=i, j=j+1
            minDistBacktracing(i, j + 1, dist + w[i][j], w, n);
        }
    }
}