108 lines
3.9 KiB
Java
108 lines
3.9 KiB
Java
/*
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* The MIT License (MIT)
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*
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* Copyright © 2022 xrv <xrv@live.com>
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*
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* Permission is hereby granted, free of charge, to any person obtaining a copy
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* of this software and associated documentation files (the "Software"), to deal
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* in the Software without restriction, including without limitation the rights
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* to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
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* copies of the Software, and to permit persons to whom the Software is
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* furnished to do so, subject to the following conditions:
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*
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* The above copyright notice and this permission notice shall be included in
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* all copies or substantial portions of the Software.
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*
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* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
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* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
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* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
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* AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
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* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
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* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
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* THE SOFTWARE.
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*/
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package io.github.ehlxr.algorithm.dp;
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/**
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* 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。
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* 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。
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* 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?
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*
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* @author ehlxr
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* @since 2022-03-05 16:46.
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*/
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public class MinDist {
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private final int[][] matrix = {{1, 3, 5, 9}, {2, 1, 3, 4}, {5, 2, 6, 7}, {6, 8, 4, 3}};
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private final int[][] mem = new int[4][4];
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private int minDist = Integer.MAX_VALUE; // 全局变量或者成员变量
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/**
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* 动态规划:状态转移表法
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*/
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public int minDistDP(int[][] matrix, int n) {
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int[][] states = new int[n][n];
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int sum = 0;
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for (int j = 0; j < n; ++j) { // 初始化states的第一行数据
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sum += matrix[0][j];
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states[0][j] = sum;
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}
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sum = 0;
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for (int i = 0; i < n; ++i) { // 初始化states的第一列数据
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sum += matrix[i][0];
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states[i][0] = sum;
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}
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for (int i = 1; i < n; ++i) {
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for (int j = 1; j < n; ++j) {
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states[i][j] = matrix[i][j] + Math.min(states[i][j - 1], states[i - 1][j]);
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}
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}
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return states[n - 1][n - 1];
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}
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/**
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* 动态规划:状态转移方程法
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*/
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public int minDist(int i, int j) { // 调用minDist(n-1, n-1);
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if (i == 0 && j == 0) {
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return matrix[0][0];
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}
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// 备忘录,如果计算过,即有数字存在,则直接返回
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if (mem[i][j] > 0) {
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return mem[i][j];
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}
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int minLeft = Integer.MAX_VALUE;
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if (j - 1 >= 0) {
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minLeft = minDist(i, j - 1);
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}
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int minUp = Integer.MAX_VALUE;
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if (i - 1 >= 0) {
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minUp = minDist(i - 1, j);
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}
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int currMinDist = matrix[i][j] + Math.min(minLeft, minUp);
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mem[i][j] = currMinDist;
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return currMinDist;
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}
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/**
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* 回溯方式
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*/
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public void minDistBacktracing(int i, int j, int dist, int[][] w, int n) {
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// 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
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if (i == n && j == n) {
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if (dist < minDist) {
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minDist = dist;
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}
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return;
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}
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if (i < n) { // 往下走,更新i=i+1, j=j
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minDistBacktracing(i + 1, j, dist + w[i][j], w, n);
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}
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if (j < n) { // 往右走,更新i=i, j=j+1
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minDistBacktracing(i, j + 1, dist + w[i][j], w, n);
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}
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}
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}
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