update at 2022-03-05 17:05:21 by ehlxr
parent
226c572e59
commit
3c8752be70
|
|
@ -0,0 +1,108 @@
|
|||
/*
|
||||
* 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
|
||||
* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
|
||||
* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
|
||||
* THE SOFTWARE.
|
||||
*/
|
||||
|
||||
package io.github.ehlxr.algorithm.dp;
|
||||
|
||||
/**
|
||||
* 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？
|
||||
*
|
||||
* @author ehlxr
|
||||
* @since 2022-03-05 14:31.
|
||||
*/
|
||||
public class KnapSack3 {
|
||||
private final int[] weight = {2, 2, 4, 6, 3}; // 物品的重量
|
||||
private final int[] value = {3, 4, 8, 9, 6}; // 物品的价值
|
||||
private final int n = 5; // 物品个数
|
||||
private final int w = 9; // 背包承受的最大重量
|
||||
private int maxV = Integer.MIN_VALUE; // 结果放到 maxV 中
|
||||
|
||||
/**
|
||||
* 动态规划方式
|
||||
*
|
||||
* @param weight 物品重量
|
||||
* @param value 物品的价值
|
||||
* @param n: 物品个数
|
||||
* @param w: 背包可承载重量
|
||||
* @return 最大价值
|
||||
*/
|
||||
public static int knapsack3(int[] weight, int[] value, int n, int w) {
|
||||
int[][] states = new int[n][w + 1];
|
||||
for (int i = 0; i < n; ++i) { // 初始化 states,默认 -1
|
||||
for (int j = 0; j < w + 1; ++j) {
|
||||
states[i][j] = -1;
|
||||
}
|
||||
}
|
||||
states[0][0] = 0;
|
||||
if (weight[0] <= w) {
|
||||
states[0][weight[0]] = value[0];
|
||||
}
|
||||
for (int i = 1; i < n; ++i) { // 动态规划,状态转移
|
||||
for (int j = 0; j <= w; ++j) { // 不选择第 i 个物品
|
||||
if (states[i - 1][j] >= 0) {
|
||||
// 复制上一层状态
|
||||
states[i][j] = states[i - 1][j];
|
||||
}
|
||||
}
|
||||
// 下面的 j 表示重量
|
||||
for (int j = 0; j <= w - weight[i]; ++j) { // 选择第 i 个物品
|
||||
if (states[i - 1][j] >= 0) { // 表示上一层存在重量为 j 的状态
|
||||
int v = states[i - 1][j] + value[i]; // 上一层重量为 j 的价值 + i 物品的价值
|
||||
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
|
||||
states[i][j + weight[i]] = v;
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
// 找出最大值
|
||||
int maxvalue = -1;
|
||||
for (int j = 0; j <= w; ++j) {
|
||||
if (states[n - 1][j] > maxvalue) {
|
||||
maxvalue = states[n - 1][j];
|
||||
}
|
||||
}
|
||||
return maxvalue;
|
||||
}
|
||||
|
||||
/**
|
||||
* 回溯方式
|
||||
*
|
||||
* @param i 第几个物品
|
||||
* @param cw 物品的重量
|
||||
* @param cv 物品的价值
|
||||
*/
|
||||
public void f(int i, int cw, int cv) { // 调用 f (0, 0, 0)
|
||||
if (cw == w || i == n) { //cw==w 表示装满了,i==n 表示物品都考察完了
|
||||
if (cv > maxV) {
|
||||
maxV = cv;
|
||||
}
|
||||
return;
|
||||
}
|
||||
f(i + 1, cw, cv); // 选择不装第 i 个物品
|
||||
if (cw + weight[i] <= w) {
|
||||
f(i + 1, cw + weight[i], cv + value[i]); // 选择装第 i 个物品
|
||||
}
|
||||
}
|
||||
|
||||
|
||||
}
|
||||
|
|
@ -0,0 +1,107 @@
|
|||
/*
|
||||
* 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
|
||||
* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
|
||||
* 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
|
||||
|
||||
/**
|
||||
* 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];
|
||||
}
|
||||
|
||||
/**
|
||||
* 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);
|
||||
}
|
||||
}
|
||||
}
|
||||
Loading…
Reference in New Issue