Java-POJ1010-STAMP

Posted 2020-11-25 Jason-Cow

说良心话，题目不难，但是题目真的很不好懂，解读一下吧

题意：

　　读入分两行，第一行为邮票面额（面额相同也视为种类不同）以0结束，第二行为顾客要求的面额，以0结束

　　要求：每个顾客最多拿4张邮票，并求最优解

　　输出：对于每个顾客要求输出一行

对于最优解的定义：

　　1.要求邮票种类尽量多（原则上每种邮票可以无限供应）

　　2.满足条件1的情况下，要求顾客拿到的邮票张数尽量少

　　3.满足条件2的情况下，要求顾客拿到的邮票中，面额最大的邮票面额越大越好

　　4.若满足以上3个条件的解不唯一，输出该顾客拿到的邮票种类数，以及“tie”

　　5.无法满足顾客要求输出“ ---- none”

PS：（聪明人才能看到我给的提示）

　　bug1，由于poj没有specail judge，面额的拼凑方案必须按从小到大排序

　　bug2，邮票种类远不止题目中给的25种，数组开小了会RE,建议开大点，第一次开了60都不够，坑爹啊！

 

 

 1 package poj.ProblemSet;
 2 
 3 import java.util.Scanner;
 4 
 5 public class poj1010 {
 6     
 7     public static final int MAXN = 100;
 8     public static int[] stamp_time = new int[MAXN];
 9     public static int[] stamp = new int[MAXN];
10     public static int[] customer = new int[MAXN];
11     public static int[] ans = new int[MAXN];
12     public static int stamp_n, customer_n, flag;
13     public static int ans_type, ans_total_time, ans_max_stamp;
14     public static int now_type, now_total_time, now_max_stamp;
15 
16     public static void query() {
17         now_type = now_total_time = now_max_stamp = 0;
18         for (int i = 1; i <= stamp_n; i++) {
19             if (stamp_time[i] > 0) {
20                 now_type++;
21                 now_total_time += stamp_time[i];
22                 if (stamp[i] > now_max_stamp) 
23                     now_max_stamp = stamp[i];
24             }
25         }
26     }
27 
28     public static void update(int type) {
29         ans_type = type;
30         ans_total_time = ans_max_stamp = 0;
31         for (int i = 1; i <= stamp_n; i++) {
32             ans[i] = stamp_time[i];
33             if (ans[i] > 0) {
34                 ans_total_time += ans[i];
35                 if (stamp[i] > ans_max_stamp) 
36                     ans_max_stamp = stamp[i];
37             }
38         }
39     }
40 
41     public static void dfs(int cost, int type, int a, int num) {
42         if (num > 4 || cost < 0) return;
43         if (cost == 0) {
44             if (type > ans_type) { flag = 0;update(type); } 
45             else if (type == ans_type) {
46                 query();
47                 if (now_total_time < ans_total_time) { flag = 0;update(type); } 
48                 else if (now_total_time == ans_total_time) {
49                     if (now_max_stamp > ans_max_stamp) { flag = 0;update(type); } 
50                     else if (now_max_stamp == ans_max_stamp) flag = 1;
51                 }
52             }
53 
54         }
55         for (int i = a; i <= stamp_n; i++) {
56             if (i == a) {
57                 if (type == 0) { stamp_time[a]++;dfs(cost - stamp[a], 1, a, num + 1);stamp_time[a]--; } 
58                 else { stamp_time[a]++;dfs(cost - stamp[a], type, a, num + 1);stamp_time[a]--; }
59             } 
60             else { stamp_time[i]++;dfs(cost - stamp[i], type + 1, i, num + 1);stamp_time[i]--; }
61         }
62     }
63 
64     public static void main(String[] args) {
65         Scanner cin = new Scanner(System.in);
66         while (cin.hasNext()) {
67             stamp_n = customer_n = 0;
68             for (int i = 0; i < MAXN; i++) stamp_time[i] = stamp[i] = customer[i] = 0;
69             do stamp[++stamp_n] = cin.nextInt(); while (stamp[stamp_n] != 0);
70             do customer[++customer_n] = cin.nextInt(); while (customer[customer_n] != 0);
71             stamp_n--;
72             customer_n--;
73             for (int i = 1; i <= stamp_n; i++)
74                 for (int j = i + 1; j <= stamp_n; j++)
75                     if (stamp[i] > stamp[j]) {
76                         int temp = stamp[i];
77                         stamp[i] = stamp[j];
78                         stamp[j] = temp;
79                     }
80             for (int i = 1; i <= customer_n; i++) {
81                 ans_type = ans_total_time = ans_max_stamp = flag = 0;
82                 for (int j = 0; j < MAXN; j++) ans[j] = 0;
83                 dfs(customer[i], 0, 1, 0);
84                 if (flag == 1) System.out.println(customer[i] + " (" + ans_type + "): tie");
85                 else if (ans_type > 0) {
86                     System.out.print(customer[i] + " (" + ans_type + "):");
87                     for (int j = 1; j <= stamp_n; j++) if (ans[j] > 0) while (ans[j]-- > 0) System.out.print(" " + stamp[j]);
88                     System.out.println();
89                 } 
90                 else System.out.println(customer[i] + " ---- none");
91             }
92         }
93     }
94 }