• Java-POJ1014-Dividing


    多重背包问题的特点是物品数量可以大于1但是有限制。状态定义与01背包一致。

    多重背包的解法有多种,复杂度也各不相同。

    对于物品数Ci较大的数据,可以采取二进制数进行优化(就是这样,别问就是baidu!)

    如何理解多重背包的二进制优化?

    使得每种物品的转移次数由O(MxCi)变为O(Mxlog(Ci))

    f[i]=f[i] | f[i-a[i]] | …| f[i-a[i]xc[i]](a[i]为物品价值,c[i]为物品数量)

    二进制优化后令W[i]=a[i]xc[i](c[i]被二进制分拆)

    /*
    Memory: 3508K        Time: 329MS
    Language: Java        Result: Accepted
    */

    实现见代码:

     1 package poj.ProblemSet;
     2 
     3 import java.util.Scanner;
     4 
     5 public class poj1014 {
     6     public static final int MAXN = 400000;
     7     public static boolean[] f = new boolean[MAXN];
     8     public static int[] w = new int[100];
     9 
    10     public static void main(String[] args) {
    11         Scanner cin = new Scanner(System.in);
    12         for (int Case = 0; cin.hasNext(); ) {
    13             int value = 0, cnt = 0;
    14             boolean flag = false;
    15             for (int i = 1; i <= 6; i++) {
    16                 int val = cin.nextInt();
    17                 value += val * i;
    18                 for (int j = 1, x = 0; val > 0; j *= 2) {
    19                     x = Math.min(j, val);
    20                     w[++cnt] = i * x;
    21                     val -= x;
    22                 }
    23             }
    24             if (value == 0) break;
    25             System.out.println("Collection #" + (++Case) + ":");
    26             if (value % 2 == 0) {
    27                 f[0] = true;
    28                 for (int i = 1; i < MAXN; i++) f[i] = false;
    29                 for (int i = 1; i <= cnt; i++)
    30                     for (int j = value / 2; j >= w[i]; j--)
    31                         f[j] |= f[j - w[i]];
    32                 flag = true;
    33             }
    34             System.out.println((!flag?"Can't":(f[value/2]?"Can":"Can't"))+" be divided.");
    35             System.out.println();
    36         }
    37     }
    38 }

     PS:另外一种O(VN)的方法是用数据结构单调队列优化!!!Orz

    /*
    Memory: 6064K        Time: 282MS
    Language: Java        Result: Accepted
    */
     1 package poj.ProblemSet;
     2 
     3 import java.util.Scanner;
     4 
     5 public class poj1014 {
     6     public static final int MAXN = 400000;
     7     public static int[] queue = new int[MAXN];
     8     public static boolean[] f = new boolean[MAXN];
     9     public static int[] c = new int[7];
    10     public static void main(String[] args) {
    11         Scanner cin = new Scanner(System.in);
    12         for (int Case = 0; cin.hasNext(); ) {
    13             int value = 0;
    14             boolean flag = false;
    15             for (int i = 1; i <= 6; i++) {
    16                 c[i] = cin.nextInt();
    17                 value += i * c[i];
    18             }
    19             if (value == 0) break;
    20             System.out.println("Collection #" + (++Case) + ":");
    21             if (value % 2 == 0) {
    22                 f[0] = true;
    23                 for (int i = 1; i <= value / 2; i++) f[i] = false;
    24                 for (int i = 1, x = 0; i <= 6; i++) {
    25                     x = i * c[i];
    26                     for (int j = 0; j < i; j++)
    27                         if (f[j]) queue[j] = j;
    28                         else queue[j] = -MAXN;
    29                     for (int j = i; j <= value / 2; j++)
    30                         if (f[j]) queue[j] = j;
    31                         else {
    32                             queue[j] = queue[j - i];
    33                             if (queue[j - i] + x >= j) f[j] = true;
    34                         }
    35                 }
    36                 flag = true;
    37             }
    38             System.out.println((!flag?"Can't":(f[value/2]?"Can":"Can't"))+" be divided.");
    39             System.out.println();
    40         }
    41     }
    42 }
    ~~Jason_liu O(∩_∩)O
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  • 原文地址:https://www.cnblogs.com/JasonCow/p/12248400.html
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