USACO Section 3.1: Stamps

这题一开始用了dfs（注释部分），结果TLE，后来想了DP方法，f[i] = f[j] + f[i-j]， j = 1, 2... i/2， 还是TLE，网上搜了别人的代码，发现自己的状态方程有问题，应该是f[i] = f[i-stamp[j]]+1， j = 1...N。这样j从1到N的话复杂度大大降低了。

 1 /*

 2 ID: yingzho1

 3 LANG: C++

 4 TASK: stamps

 5 */

 6 #include <iostream>

 7 #include <fstream>

 8 #include <string>

 9 #include <map>

10 #include <vector>

11 #include <set>

12 #include <algorithm>

13 #include <stdio.h>

14 #include <queue>

15 #include <cstring>

16 #include <cmath>

17 #include <list>

18 #include <cstdio>

19 #include <cstdlib>

20 

21 using namespace std;

22 

23 ifstream fin("stamps.in");

24 ofstream fout("stamps.out");

25 

26 const int inf = 100000000;

27 

28 int K, N;

29 

30 /*bool check(int cur, vector<int> &stamp, int total, int dep) {

31     if (cur < 0 || total < 0) return false;

32     if (cur == 0) return true;

33     for (int i = dep; i < stamp.size(); i++) {

34         if (check(cur-stamp[i], stamp, total-1, i)) return true;

35     }

36     return false;

37 }

38 

39 bool cmp(const int a, const int b) {return a > b;}*/

40 

41 int main()

42 {

43     fin >> K >> N;

44     vector<int> stamp(N);

45     set<int> S;

46     for (int i = 0; i < N; i++) {

47         fin >> stamp[i];

48         S.insert(stamp[i]);

49     }

50 //    sort(stamp.begin(), stamp.end(), cmp);

51 /*    if (stamp[stamp.size()-1] != 1) {

52         fout << 0 << endl;

53         return 0;

54     }*/

55     vector<int> f(1);

56     int cur = 1;

57     while (cur) {

58         //cout << cur << endl;

59         if (S.count(cur)) f.push_back(1);

60         else {

61             int tmp = inf;

62             //for (int i = 1; i <= cur/2; i++) tmp = min(tmp, f[i]+f[cur-i]);

63             for (int i = 0; i < N; i++) {

64                 if (cur > stamp[i]) tmp = min(tmp, f[cur-stamp[i]]+1);

65             }

66             if (tmp > K) break;

67             f.push_back(tmp);

68         }

69         cur++;

70     }

71     fout << cur-1 << endl;

72 

73 

74     /*int cur = 2;

75     while (check(cur, stamp, K, 0)) {

76         //cout << cur << endl;

77         cur++;

78     }

79     fout << cur-1 << endl;*/

80 

81     return 0;

82 }