DP还有很长很长一段路要走。。
题意：给出n纪念品的价格和钱数m，问最多能买多少件纪念品和买这些数量的纪念品的方案数。
首先，求能买最多的纪念品的数量，用贪心法可以解决。将价钱排序，然后从最便宜的开始买，这样就很容易求得最多买的纪念品的数量。
方案数就要用到动态规划。
dp[j][k]表示花费不超过j元买k件物品的方案数
dp[j][k] += dp[j-a[i]][k-1]
因为这里本来是个三维数组的，多一个维度用来表示前i件物品。调整了循环顺序，类似01背包空间上的优化，所以倒着循环就可以利用之前的计算结果节省空间。
1 //#define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <algorithm> 6 using namespace std; 7 8 int dp[8][4], a[505]; 9 10 int main(void) 11 { 12 #ifdef LOCAL 13 freopen("2126in.txt", "r", stdin); 14 #endif 15 16 int T; 17 scanf("%d", &T); 18 while(T--) 19 { 20 int n, m, num = 0, sum = 0; 21 scanf("%d%d", &n, &m); 22 for(int i = 1; i <= n; ++i) 23 scanf("%d", &a[i]); 24 25 sort(a+1, a+1+n); 26 for(int i = 1; i <= n; ++i) 27 { 28 sum += a[i]; 29 if(m >= sum) 30 ++num; 31 } 32 if(num == n) 33 { 34 printf("You have 1 selection(s) to buy with %d kind(s) of souvenirs. ", num); 35 continue; 36 } 37 if(num == 0) 38 { 39 printf("Sorry, you can't buy anything. "); 40 continue; 41 } 42 memset(dp, 0, sizeof(dp)); 43 for(int i = 0; i <= m; ++i) 44 dp[i][0] = 1; 45 for(int i = 1; i <= n; ++i) 46 for(int j = m; j >= a[i]; --j) 47 for(int k = num; k >= 1; --k) 48 dp[j][k] = dp[j][k] + dp[j-a[i]][k-1]; 49 50 if(dp[m][num] == 0) 51 printf("Sorry, you can't buy anything. "); 52 else 53 printf("You have %d selection(s) to buy with %d kind(s) of souvenirs. ", dp[m][num], num); 54 } 55 return 0; 56 }
解法二：
看注释吧。。
1 //#define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 7 int a[35]; 8 int dp[35][505][2]; //dp[i][j][0]表示前i件物品花费不超过j元 9 //能最多买的礼物数，dp[i][j][1]表示方案数 10 11 int main(void) 12 { 13 #ifdef LOCAL 14 freopen("2126in.txt", "r", stdin); 15 #endif 16 17 int T; 18 scanf("%d", &T); 19 while(T--) 20 { 21 int n, m; 22 scanf("%d%d", &n, &m); 23 for(int i = 1; i <= n; ++i) 24 scanf("%d", &a[i]); 25 memset(dp, 0, sizeof(dp)); 26 //边界初始化 27 for(int i = 0; i <= n; ++i) 28 dp[i][0][1] = 1; 29 for(int j = 0; j <= m; ++j) 30 dp[0][j][1] = 1; 31 for(int i = 1; i <= n; ++i) 32 { 33 for(int j = 0; j <= m; ++j) 34 { 35 if(j < a[i] || dp[i-1][j-a[i]][0]+1 < dp[i-1][j][0]) 36 {//如果买第i件礼物礼物总数变少，那么肯定不买! 37 dp[i][j][0] = dp[i-1][j][0]; 38 dp[i][j][1] = dp[i-1][j][1]; 39 } 40 else if(dp[i-1][j-a[i]][0] + 1 == dp[i-1][j][0]) 41 {//买的礼物数相同，则方案数为买和不买第i件礼物的方案数的总和 42 dp[i][j][0] = dp[i-1][j][0]; 43 dp[i][j][1] = dp[i-1][j][1] + dp[i-1][j-a[i]][1]; 44 } 45 else //dp[i-1][j-a[i]][0] + 1 > dp[i-1][j][0] 46 {//可以买更多数量的礼物 47 dp[i][j][0] = dp[i-1][j-a[i]][0] + 1; 48 dp[i][j][1] = dp[i-1][j-a[i]][1]; 49 } 50 } 51 } 52 53 if(dp[n][m][0]) 54 printf("You have %d selection(s) to buy with %d kind(s) of souvenirs. ", dp[n][m][1], dp[n][m][0]); 55 else 56 printf("Sorry, you can't buy anything. "); 57 } 58 return 0; 59 }
解法二空间上的优化之滚动数组：
对了，那个else不能去掉！血一般的教训
1 #define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 using namespace std; 6 7 int a[35]; 8 int dp[505][2]; 9 10 int main(void) 11 { 12 #ifdef LOCAL 13 freopen("2126in.txt", "r", stdin); 14 #endif 15 16 int T; 17 scanf("%d", &T); 18 while(T--) 19 { 20 int n, m; 21 scanf("%d%d", &n, &m); 22 for(int i = 1; i <= n; ++i) 23 scanf("%d", &a[i]); 24 memset(dp, 0, sizeof(dp)); 25 for(int i = 0; i <= m; ++i) //处理边界 26 dp[i][1] = 1; 27 for(int i = 1; i <= n; ++i) 28 { 29 dp[0][1] = 1; 30 for(int j = m; j >= a[i]; --j) 31 { 32 if(dp[j-a[i]][0] + 1 > dp[j][0]) 33 { 34 dp[j][0] = dp[j-a[i]][0] + 1; 35 dp[j][1] = dp[j-a[i]][1]; 36 } 37 else if(dp[j-a[i]][0] + 1 == dp[j][0]) 38 { 39 dp[j][1] = dp[j][1] + dp[j-a[i]][1]; 40 } 41 } 42 } 43 if(dp[m][0]) 44 printf("You have %d selection(s) to buy with %d kind(s) of souvenirs. ", dp[m][1], dp[m][0]); 45 else 46 printf("Sorry, you can't buy anything. "); 47 } 48 return 0; 49 }
最后吐槽一下我发现的一个奇怪的现象：如果源代码的文件名里带括号，这样下断点调试的时候会自动退出的。CFree-5和DevCpp都是这样的，=_=||