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。

 /*
 ID: yingzho1
 LANG: C++
 TASK: stamps
 */
 #include <iostream>
 #include <fstream>
 #include <string>
 #include <map>
 #include <vector>
 #include <set>
 #include <algorithm>
 #include <stdio.h>
 #include <queue>
 #include <cstring>
 #include <cmath>
 #include <list>
 #include <cstdio>
 #include <cstdlib>

 using namespace std;

 ifstream fin("stamps.in");
 ofstream fout("stamps.out");

 ;

 int K, N;

 /*bool check(int cur, vector<int> &stamp, int total, int dep) {
     if (cur < 0 || total < 0) return false;
     if (cur == 0) return true;
     for (int i = dep; i < stamp.size(); i++) {
         if (check(cur-stamp[i], stamp, total-1, i)) return true;
     }
     return false;
 }

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

 int main()
 {
     fin >> K >> N;
     vector<int> stamp(N);
     set<int> S;
     ; i < N; i++) {
         fin >> stamp[i];
         S.insert(stamp[i]);
     }
 //    sort(stamp.begin(), stamp.end(), cmp);
 /*    if (stamp[stamp.size()-1] != 1) {
         fout << 0 << endl;
         return 0;
     }*/
     vector<);
     ;
     while (cur) {
         //cout << cur << endl;
         );
         else {
             int tmp = inf;
             //for (int i = 1; i <= cur/2; i++) tmp = min(tmp, f[i]+f[cur-i]);
             ; i < N; i++) {
                 );
             }
             if (tmp > K) break;
             f.push_back(tmp);
         }
         cur++;
     }
     fout << cur- << endl;

     /*int cur = 2;
     while (check(cur, stamp, K, 0)) {
         //cout << cur << endl;
         cur++;
     }
     fout << cur-1 << endl;*/

     ;
 }