1068. Find More Coins (30)

动态背包

题意：

给定一系列的硬币值, 然后给定一个目标value， 从所有硬币中找出几个, 使得这几个硬币的和正好等于这个value, 而且这个硬币序列应该是满足硬币值字典序的最小序列.

分析：

属于典型的背包问题. 用动态规划(dp)做, 假设F(N, M)表示不超过面值M, 而且从前面N个硬币中挑选硬币值能得到的最大硬币面值总和, 那我们可以得到如下递归公式：

F(N, M) = max{ F(N–1, M), F(N–1, M–c(N)) + c(N) }，c(N)表示第N个硬币的面值

那么最后要是F(N, M) == M， 那么就说明我们可以找到这样一组硬币, 使得他们的面值总和恰好等于M。

记录路径： 若has(N, M)为真, 则表示从前面N个硬币中选出一组得到最多的不超过M的币值总和里面包括了第N个硬币。那么如果我们需要找到完整路径, 我们就可以从N、M出发，一直回溯到最后一个硬币。

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she must pay the exact amount. Since she has as many as 10⁴ coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find some coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (<=10⁴, the total number of coins) and M(<=10², the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the face values V₁ <= V₂ <= ... <= V_k such that V₁ + V₂ + ... + V_k = M. All the numbers must be separated by a space, and there must be no extra space at the end of the line. If such a solution is not unique, output the smallest sequence. If there is no solution, output "No Solution" instead.

Note: sequence {A[1], A[2], ...} is said to be "smaller" than sequence {B[1], B[2], ...} if there exists k >= 1 such that A[i]=B[i] for all i < k, and A[k] < B[k].

Sample Input 1:

8 9
5 9 8 7 2 3 4 1

Sample Output 1:

1 3 5

Sample Input 2:

4 8
7 2 4 3

Sample Output 2:

No Solution

来源： <http://www.patest.cn/contests/pat-a-practise/1068>

 

1. #include<stdio.h>
2. #include<iostream>
3. #include<string>
4. #include<string.h>
5. #include<algorithm>
6. #include<vector>
7. #pragma warning(disable:4996)
9. using namespace std;
11. #define MAXTOTAL 10001
12. #define MAXAMOUNT 101
14. int f[MAXTOTAL][MAXAMOUNT];//f[n][m]表示 前n个数中 得出的 最接近m 的值
15. bool has[MAXTOTAL][MAXAMOUNT];//has[n][m]表示在前n个数中得出最接近m的值时 是否用到c[n]
16. int *c = NULL;
18. int CalcClosestSum(int n, int m) {
19. memset(f,0,sizeof(f));
20. memset(has, false, sizeof(has));
21. int i, j;
22. int sec;
23. for (int i = 1; i <= n; i++) {
24. for (int j = 1; j <= m; j++) {
25. if (j - c[i] < 0)
26. sec = 0;
27. else
28. sec = f[i - 1][j - c[i]] + c[i];
29. if (f[i - 1][j]>sec)
30. f[i][j] = f[i - 1][j];
31. else {
32. f[i][j] = sec;
33. has[i][j] = true;
34. }
35. }
36. }
37. return f[n][m];
38. }
40. bool cmp(const int& a, const int& b) {
41. return a > b;
42. }
44. int main(void) {
45. freopen("Text.txt", "r", stdin);
46. int n, m;
47. cin >> n >> m;
48. c = new int[n + 1];
49. memset(c, 0, sizeof(int)*n + 1);
51. int i;
52. for (int i = 0; i < n; i++) {
53. cin >> c[i + 1];
54. }
55. sort(&c[1], &c[n + 1], cmp);
57. int res = CalcClosestSum(n, m);
58. if (res == m) {
59. vector<int> v;
60. while (m)
61. {
62. while (!has[n][m])
63. {
64. n--;
65. }
66. v.push_back(c[n]);
67. m = m - c[n];
68. n--;
69. }
70. for ( i = 0; i < v.size() - 1; i++)
71. cout << v[i] << " ";
72. cout << v[i] << endl;
73. }
74. else
75. cout << "No Solution" << endl;
77. return 0;
78. }

来自为知笔记(Wiz)

转载于:https://www.cnblogs.com/zzandliz/p/5023213.html