vijos 1037 ***

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 1 #include <cstdio>

 2 #include <cstring>

 3 #include <algorithm>

 4 #include <iostream>

 5 using namespace std;

 6 const int v = 2000 + 5;

 7 const int MaxN = 100 + 5;

 8 int N, sum, num[MaxN], dp[2][v];

 9 int main() {

10     int i, j, k;

11     cin >> N;

12     for(i = 0; i < N; ++i) {

13         cin >> num[i];

14         sum += num[i];

15     }

16     memset(dp, -1, sizeof(dp));

17     dp[1][0] = 0;

18     int a;

19     for(i = 0; i < N; ++i) {

20         a = i % 2;

21         memset(dp[a], -1, sizeof(dp[a]));

22         for(j = 0; j <= sum; ++j) {

23             //1.不放第i块水晶;

24             if(dp[a ^ 1][j] > -1)

25                 dp[a][j] = dp[a ^ 1][j];

26             //2.放进去后，高塔变矮塔（第i块放在矮塔上了）;

27             if(num[i] > j &&  dp[a ^ 1][num[i] - j] > -1)

28                 dp[a][j] = max(dp[a][j], dp[a ^ 1][num[i] - j] + j);

29             //3.放进去后，高塔仍高（第i块放在矮塔上）;

30             if(j + num[i] <= sum && dp[a ^ 1][j + num[i]] > -1)

31                 dp[a][j] = max(dp[a][j], dp[a ^ 1][j + num[i]]);

32             //4.放进去后，高塔更高（第i块放在高塔上）.

33             if(j >= num[i] && dp[a ^ 1][j - num[i]] > -1)

34                 dp[a][j] = max(dp[a][j], dp[a ^ 1][j - num[i]] + num[i]);

35         }

36     }

37     if(dp[a][0] > 0)

38         cout << dp[a][0] << endl;

39     else

40         cout << "Impossible" << endl;

41 }