update at 2022-03-05 17:05:21 by ehlxr
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/*
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* The MIT License (MIT)
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*
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* Copyright © 2022 xrv <xrg@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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* 对于一组不同重量、不同价值、不可分割的物品，我们选择将某些物品装入背包，在满足背包最大重量限制的前提下，背包中可装入物品的总价值最大是多少呢？
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*
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* @author ehlxr
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* @since 2022-03-05 14:31.
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*/
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public class KnapSack3 {
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private final int[] weight = {2, 2, 4, 6, 3}; // 物品的重量
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private final int[] value = {3, 4, 8, 9, 6}; // 物品的价值
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private final int n = 5; // 物品个数
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private final int w = 9; // 背包承受的最大重量
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private int maxV = Integer.MIN_VALUE; // 结果放到 maxV 中
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/**
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* 动态规划方式
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*
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* @param weight 物品重量
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* @param value 物品的价值
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* @param n: 物品个数
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* @param w: 背包可承载重量
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* @return 最大价值
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*/
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public static int knapsack3(int[] weight, int[] value, int n, int w) {
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int[][] states = new int[n][w + 1];
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for (int i = 0; i < n; ++i) { // 初始化 states,默认 -1
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for (int j = 0; j < w + 1; ++j) {
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states[i][j] = -1;
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}
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}
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states[0][0] = 0;
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if (weight[0] <= w) {
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states[0][weight[0]] = value[0];
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}
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for (int i = 1; i < n; ++i) { // 动态规划,状态转移
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for (int j = 0; j <= w; ++j) { // 不选择第 i 个物品
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if (states[i - 1][j] >= 0) {
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// 复制上一层状态
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states[i][j] = states[i - 1][j];
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}
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}
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// 下面的 j 表示重量
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for (int j = 0; j <= w - weight[i]; ++j) { // 选择第 i 个物品
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if (states[i - 1][j] >= 0) { // 表示上一层存在重量为 j 的状态
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int v = states[i - 1][j] + value[i]; // 上一层重量为 j 的价值 + i 物品的价值
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if (v > states[i][j + weight[i]]) { // states[i][j + weight[i]]：存储当前重量 j + 物品 i 的重量所在位置的价值
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states[i][j + weight[i]] = v;
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}
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}
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}
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}
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// 找出最大值
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int maxvalue = -1;
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for (int j = 0; j <= w; ++j) {
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if (states[n - 1][j] > maxvalue) {
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maxvalue = states[n - 1][j];
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}
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}
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return maxvalue;
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}
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/**
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* 回溯方式
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*
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* @param i 第几个物品
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* @param cw 物品的重量
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* @param cv 物品的价值
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*/
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public void f(int i, int cw, int cv) { // 调用 f (0, 0, 0)
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if (cw == w || i == n) { //cw==w 表示装满了,i==n 表示物品都考察完了
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if (cv > maxV) {
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maxV = cv;
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}
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return;
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}
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f(i + 1, cw, cv); // 选择不装第 i 个物品
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if (cw + weight[i] <= w) {
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f(i + 1, cw + weight[i], cv + value[i]); // 选择装第 i 个物品
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}
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}
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}
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@@ -0,0 +1,107 @@
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/*
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* The MIT License (MIT)
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*
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* Copyright © 2022 xrv <xrg@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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* 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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* 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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