Java-POJ1010-STAMP

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！

 package poj.ProblemSet;

 import java.util.Scanner;

 public class poj1010 {

     public static final int MAXN = 100;
     public static int[] stamp_time = new int[MAXN];
     public static int[] stamp = new int[MAXN];
     public static int[] customer = new int[MAXN];
     public static int[] ans = new int[MAXN];
     public static int stamp_n, customer_n, flag;
     public static int ans_type, ans_total_time, ans_max_stamp;
     public static int now_type, now_total_time, now_max_stamp;

     public static void query() {
         now_type = now_total_time = now_max_stamp = 0;
         for (int i = 1; i <= stamp_n; i++) {
             if (stamp_time[i] > 0) {
                 now_type++;
                 now_total_time += stamp_time[i];
                 if (stamp[i] > now_max_stamp)
                     now_max_stamp = stamp[i];
             }
         }
     }

     public static void update(int type) {
         ans_type = type;
         ans_total_time = ans_max_stamp = 0;
         for (int i = 1; i <= stamp_n; i++) {
             ans[i] = stamp_time[i];
             if (ans[i] > 0) {
                 ans_total_time += ans[i];
                 if (stamp[i] > ans_max_stamp)
                     ans_max_stamp = stamp[i];
             }
         }
     }

     public static void dfs(int cost, int type, int a, int num) {
         if (num > 4 || cost < 0) return;
         if (cost == 0) {
             if (type > ans_type) { flag = 0;update(type); }
             else if (type == ans_type) {
                 query();
                 if (now_total_time < ans_total_time) { flag = 0;update(type); }
                 else if (now_total_time == ans_total_time) {
                     if (now_max_stamp > ans_max_stamp) { flag = 0;update(type); }
                     else if (now_max_stamp == ans_max_stamp) flag = 1;
                 }
             }

         }
         for (int i = a; i <= stamp_n; i++) {
             if (i == a) {
                 if (type == 0) { stamp_time[a]++;dfs(cost - stamp[a], 1, a, num + 1);stamp_time[a]--; }
                 else { stamp_time[a]++;dfs(cost - stamp[a], type, a, num + 1);stamp_time[a]--; }
             }
             else { stamp_time[i]++;dfs(cost - stamp[i], type + 1, i, num + 1);stamp_time[i]--; }
         }
     }

     public static void main(String[] args) {
         Scanner cin = new Scanner(System.in);
         while (cin.hasNext()) {
             stamp_n = customer_n = 0;
             for (int i = 0; i < MAXN; i++) stamp_time[i] = stamp[i] = customer[i] = 0;
             do stamp[++stamp_n] = cin.nextInt(); while (stamp[stamp_n] != 0);
             do customer[++customer_n] = cin.nextInt(); while (customer[customer_n] != 0);
             stamp_n--;
             customer_n--;
             for (int i = 1; i <= stamp_n; i++)
                 for (int j = i + 1; j <= stamp_n; j++)
                     if (stamp[i] > stamp[j]) {
                         int temp = stamp[i];
                         stamp[i] = stamp[j];
                         stamp[j] = temp;
                     }
             for (int i = 1; i <= customer_n; i++) {
                 ans_type = ans_total_time = ans_max_stamp = flag = 0;
                 for (int j = 0; j < MAXN; j++) ans[j] = 0;
                 dfs(customer[i], 0, 1, 0);
                 if (flag == 1) System.out.println(customer[i] + " (" + ans_type + "): tie");
                 else if (ans_type > 0) {
                     System.out.print(customer[i] + " (" + ans_type + "):");
                     for (int j = 1; j <= stamp_n; j++) if (ans[j] > 0) while (ans[j]-- > 0) System.out.print(" " + stamp[j]);
                     System.out.println();
                 }
                 else System.out.println(customer[i] + " ---- none");
             }
         }
     }
 }