dp背包3步曲
1.确定dp[i] [v]的含义(一维的话是dp[v]) :在 0…i 的物品中,体积为 v 的背包中,能够拿到的最大价值为 dp[i] [v]。
2.求关系式不拿物品:(物品数量减少)一维:dp[v] 二维:dp[i] [v] = dp[i-1] [v]拿:(物品数量减少,背包体积减物品体积)一维:dp[v-weight[i]] 二维:dp[i] [v] = dp[i-1] [v-weight[i]]一维背包:
//遍历物品的数量Nfor (int i = 1; i <= n; i++) {//遍历背包的容量V,但是从大到小进行遍历,如果背包容量大于物品体积就拿for (int j = capacity; j >= weights[i]; j--) {//不拿(背包体积不变)和拿(背包体积减少再加上物品价值),进行比大小,取大值。dp[j] = Math.max(dp[j], dp[j - weights[i]] + values[i]);}}
二维背包:
//遍历物品的数量Nfor(int i = 1;i<= n ; i++){//遍历背包的容量Vfor(int j = 0; j <= c ; j++ ){//如果背包体积小于物品的体积,放不下就不拿了if(j<v[i]){dp[i][j] = dp[i-1][j];}else{//不拿(物品数量-1)和拿(物品数量-1,并且背包体积减少再加上物品价值),进行比大小,取大值。dp[i][j] = Math.max(dp[i-1][j],dp[i-1][j-w[i]]+v[i]);}}}
二维背包
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int numCases = sc.nextInt(); // Number of test casesfor (int caseNum = 0; caseNum < numCases; caseNum++) {int numItems = sc.nextInt(); // Number of itemsint capacity = sc.nextInt(); // Maximum capacity of the knapsackint limit = sc.nextInt(); // Maximum weight limitint[] values = new int[numItems + 1];int[] weights = new int[numItems + 1];int[] profits = new int[numItems + 1];for (int i = 1; i <= numItems; i++) {values[i] = sc.nextInt(); // Value of the itemweights[i] = sc.nextInt(); // Weight of the itemprofits[i] = sc.nextInt(); // Profit of the item}int[][] dp = new int[capacity + 1][limit + 1]; // DP tablefor (int i = 1; i <= numItems; i++) {for (int j = capacity; j >= values[i]; j--) {for (int k = limit; k >= weights[i]; k--) {dp[j][k] = Math.max(dp[j][k], dp[j - values[i]][k - weights[i]] + profits[i]);}}}System.out.println(dp[capacity][limit]);}sc.close();}
}
如果每个物品只能用一次,反向更新是必要的;如果每个物品可以用多次,则可以正向更新。
01背包
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int numCases = sc.nextInt(); // Number of test casesfor (int caseNum = 0; caseNum < numCases; caseNum++) {int numItems = sc.nextInt(); // Number of itemsint capacity = sc.nextInt(); // Maximum capacity of the knapsackint limit = sc.nextInt(); // Maximum weight limitint[] values = new int[numItems + 1];int[] weights = new int[numItems + 1];int[] profits = new int[numItems + 1];for (int i = 1; i <= numItems; i++) {values[i] = sc.nextInt(); // Value of the itemweights[i] = sc.nextInt(); // Weight of the itemprofits[i] = sc.nextInt(); // Profit of the item}int[][] dp = new int[capacity + 1][limit + 1]; // DP tablefor (int i = 1; i <= numItems; i++) {for (int j = capacity; j >= values[i]; j--) {for (int k = limit; k >= weights[i]; k--) {dp[j][k] = Math.max(dp[j][k], dp[j - values[i]][k - weights[i]] + profits[i]);}}}System.out.println(dp[capacity][limit]);}sc.close();}
}
多重背包
public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int nt = sc.nextInt();for (int t = 0; t < nt; t++) {int n = sc.nextInt(); // 物品种类int c = sc.nextInt(); // 背包容量int[] w = new int[n + 1];int[] v = new int[n + 1];int[] m = new int[n + 1];for (int i = 1; i <= n; i++) { // 索引从1开始w[i] = sc.nextInt(); // 物品重量v[i] = sc.nextInt(); // 物品价值m[i] = sc.nextInt(); // 物品个数}int[] dp = new int[c + 1];for (int i = 1; i <= n; i++) { // 遍历物品if (m[i] * w[i] >= c) { // 如果物品数量无限多(足够多),相当于完全背包for (int j = w[i]; j <= c; j++) {dp[j] = Math.max(dp[j], dp[j - w[i]] + v[i]);}} else { // 如果物品数量有限,先用二进制优化,再当多重背包做,减少循环次数int num = m[i]; //num表示物品的剩余数量for (int k = 1; num > 0; k <<= 1) { //k表示当前二进制位的值,当前物品数量>0就说明可以进行二进制优化//取小的,确保每次取的物品数量不超过实际剩余数量,在k变大时,可能会超过num,这时就需要取实际剩余的num而不是k。int cnt = Math.min(k, num); //cnt表示取出的物品数量num -= cnt; //减去已从背包分出去的物品for (int j = c; j >= cnt * w[i]; j--) {dp[j] = Math.max(dp[j], dp[j - cnt * w[i]] + cnt * v[i]);}}}}System.out.println(dp[c]); // 输出结果}sc.close();}
}
最长公共子序列:
import java.util.Scanner;public class Main {static char[] x;static char[] y;static int[][] dp; // 字符串 X 和字符串 Y 的最长公共子序列的长度。public static void main(String[] args) {Scanner sc = new Scanner(System.in);int t = sc.nextInt();sc.nextLine(); // 吸收换行符for (int i = 1; i <= t; i++) {x = sc.nextLine().toCharArray();y = sc.nextLine().toCharArray();// 初始化dp数组,也就是x,y数组的长度int xl = x.length;int yl = y.length;dp = new int[xl + 1][yl + 1]; // 初始化dp数组大小// 循环数组从1开始,不用加dp[][],因为已经初始化了for (int j = 1; j <= xl; j++) { for (int k = 1; k <= yl; k++) {// 如果上一个字符相同,就计算出当前值,更新当前的公共子序列长度if (x[j - 1] == y[k - 1]) {dp[j][k] = dp[j - 1][k - 1] + 1;} else {// 如果不相同,在之前算出来的答案找最适合的,取公共子序列的(二维方向的)前一位进行对比,取大的作为最长公共子序列。dp[j][k] = Math.max(dp[j][k - 1], dp[j - 1][k]);}}}System.out.println(dp[xl][yl]);}sc.close(); // 关闭读写流}}
最长上升子序列:
核心思路:通过动态规划的方法,用 dp[i]
表示以 nums[i]
结尾的最长上升子序列的长度,遍历数组,更新每个位置的 dp[i]
值,同时记录整个数组的最长上升子序列长度 maxans
。
import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner scanner = new Scanner(System.in);int size = scanner.nextInt();int[] numbers = new int[size];// 读取输入数组for (int i = 0; i < size; i++) {numbers[i] = scanner.nextInt();}int result = lengthOfLIS(numbers);System.out.println(result);
}public static int lengthOfLIS(int[] nums) {// dp[i] 表示下标i在 nums[i] 时,也就是0-i的最长上升子序列的长度int[] dp = new int[nums.length];dp[0] = 1; // 初始化,每个元素自身构成一个长度为 1 的子序列int maxans = 1; // 记录最长上升子序列的长度//外层循环从1开始遍历每个元素,计算以该元素结尾的最长上升子序列的长度。for (int i = 1; i < nums.length; i++) {dp[i] = 1; // 也就是初始化dp的每一个元素为1,可以写到外面for (int j = 0; j < i; j++) {// 如果 nums[i] 大于 nums[j],则可以将 nums[i] 加入到以 nums[j] 结尾的子序列中if (nums[i] > nums[j]) {//既然要更新当前下标i位置的最长上升子序列的长度dp,那就是拿上一个已经算出来的值+1更新,当然了要进行对比取最大值,如果当前的dp[i]比dp[j]还大,那就还是使用当前的。dp[i] = Math.max(dp[i], dp[j] + 1); }}// 更新最长上升子序列的长度maxans = Math.max(maxans, dp[i]);}return maxans;
}
超市搞活动(01背包)
#include <iostream>
#include <vector>
#include <algorithm>using namespace std;int getMax(int a, int b) {if (a > b) {return a;} else {return b;}
}int knapsack(int capacity, int numItems, int profits[], int weights[]) {vector<int> dp(capacity + 1, 0);for (int i = 0; i < numItems; i++) {for (int j = weights[i]; j <= capacity; j++) {dp[j] = getMax(dp[j], dp[j - weights[i]] + profits[i]);}}return dp[capacity];
}int main() {int capacity, numItems;while (cin >> capacity >> numItems) {int profits[numItems], weights[numItems];for (int i = 0; i < numItems; i++) {cin >> profits[i] >> weights[i];}int result = knapsack(capacity, numItems, profits, weights);cout << result << endl;}return 0;
}