POJ1015 动态规划

POJ1015

问题重述：

在n个候选者中选取m个人进入陪审团。每个候选者获得两项评分：D[j]，P[j]。求解最佳评审团，使得在每个成员的两项评分和之差 最小的情况下，使得两项评分和之和 最大。

分析：

欲采用动态规划求解，必须先找到最优子结构。假如考虑评分差的绝对值，它的子问题并不一定是最优解。若考虑一定评分差下的评分和最大值，则拥有最优子结构。

用dp[i][j]表示在第i个评委评分后，评分差是j的最大评分和，得到递归公式：

dp[i][j] = max{ dp[ i - 1 ][ j - (D[i] - P[j]) ] + D[i] + P[j], dp[ i - 1 ][j] }

AC代码 

 1 //Memory: 380K        Time: 79MS

 2 #include <iostream>

 3 #include <cstring>

 4 #include <cstdio>

 5 #include <algorithm>

 6 

 7 using namespace std;

 8 

 9 int dp[21][810];

10 int prior[21][810];

11 int n, m;

12 int q[201], d[201];

13 const int offset = 405;

14 int r[21];

15 

16 bool findPath(int n, int s, int target)

17 {

18     if (n == 0) return false;

19     if (target == prior[n][s + offset]) return true;

20     int p = prior[n][s + offset];

21     return findPath(n - 1, s - (d[p] - q[p]), target);

22 }

23 

24 void output(int ans)

25 {

26     for (int j = m ; j >= 1; j--) {

27         r[j - 1] = prior[j][ans + offset];

28         int p = prior[j][ans + offset];

29         ans -= (d[p] - q[p]);

30     }

31     sort(r, r + m);

32     for (int i = 0; i < m; i++)

33         cout << " " << r[i];

34     cout << endl << endl;

35 }

36 

37 int main()

38 {

39     int cas = 1; 

40     while ( cin >> n >> m && n ) {

41         for (int i = 1; i <= n; i++)

42             cin >> d[i] >> q[i];

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

44         memset(prior, 0, sizeof(prior));

45         dp[0][offset] = 0;

46 

47         for (int i = 1; i <= m; i++) {

48             for (int j = 1; j <= n; j++) {

49                 int ss = d[j] + q[j];

50                 int dd = d[j] - q[j];

51                 for (int s = -400; s <= 400; s++) {

52                     if (s - dd < -400 || s - dd > 400) continue;

53                     if (dp[i - 1][s - dd + offset] != -1 && dp[i - 1][s - dd + offset] + ss > dp[i][s + offset] && !findPath(i - 1, s - dd, j)) {

54                         dp[i][s + offset] = dp[i - 1][s - dd + offset] + ss;

55                         prior[i][s + offset] = j;

56                     }

57                 }

58             }

59         }

60         for (int i = 0; i <= 400; i++) {

61             if (dp[m][offset + i] == -1 && dp[m][offset - i] == -1) continue;

62 

63             int ans = dp[m][offset + i] > dp[m][offset - i] ? i : -i;

64 

65             int sd = (dp[m][ans + offset] + ans) / 2;

66             int sq = (dp[m][ans + offset] - ans) / 2;

67 

68             cout << "Jury #" << cas++ << endl;

69             cout << "Best jury has value " << sd << " for prosecution and value "

70                 << sq << " for defence:" << endl;

71             output(ans);

72             break;

73         }

74     }

75     return 0;

76 }