HDU 1224 Free DIY Tour

题意：给出每个城市interesting的值，和城市之间的飞行路线，求一条闭合路线（从原点出发又回到原点）

使得路线上的interesting的值之和最大

因为要输出路径，所以用pre数组来保存前驱

在输出路径的时候，我是把前驱一次放在route数组里面，然后再将整个数组反转过来

 

另外，看别人的题解里面还有一种递归的方法求路径，挺有新意，学习了

 

我的做法：

 1 //#define LOCAL

 2 #include <iostream>

 3 #include <cstdio>

 4 #include <cstring>

 5 #include <algorithm>

 6 using namespace std;

 7 

 8 const int maxn = 5100;

 9 bool flight[110][110];

10 int city[110];

11 int dp[110];

12 int pre[110];

13 int route[110];

14 

15 int main(void)

16 {

17     #ifdef LOCAL

18         freopen("1224in.txt", "r", stdin);

19     #endif

20 

21     int N, kase;

22     scanf("%d", &N);

23     for(kase = 1; kase <= N; ++kase)

24     {

25         if(kase > 1)    printf("\n");

26 

27         int n;

28         scanf("%d", &n);

29         int i;

30         for(i = 1; i <= n; ++i)

31             scanf("%d", &city[i]);

32         city[n + 1] = 0;

33         

34         int m, from, to;

35         memset(flight, 0, sizeof(flight));

36         scanf("%d", &m);

37         for(i = 1; i <= m; ++i)

38         {

39             scanf("%d%d", &from, &to);

40             flight[from][to] = true;

41         }

42 

43         memset(dp, 0, sizeof(dp));

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

45         int j;

46         for(i = 2; i <= n + 1; ++i)

47             for(j = 1; j < i; ++j)

48                 if(flight[j][i]

49                     && (dp[j] + city[i]) > dp[i])

50                 {

51                     dp[i] = dp[j] + city[i];

52                     pre[i] = j;

53                 }

54                     

55         

56         printf("CASE %d#\n", kase);

57         printf("points : %d\n", dp[n + 1]);

58         int p = n + 1, len = 0;

59         while(pre[p] != 0)

60         {

61             route[len++] = pre[p];

62             p = pre[p];

63         }

64         for(i = 0; i < len / 2; ++i)

65         {

66             int t = route[i];

67             route[i] = route[len - i -1];

68             route[len - i -1] = t;

69         }

70         

71         printf("circuit : ");

72         for(i = 0; i < len; ++i)

73             printf("%d->", route[i]);

74         printf("1\n");

75     }

76 

77     return 0;

78 }

代码君

 

别人递归回溯输出路径的做法：

 1 //#define LOCAL

 2 #include <iostream>

 3 #include <cstdio>

 4 #include <cstring>

 5 #include <algorithm>

 6 using namespace std;

 7 

 8 const int maxn = 5100;

 9 bool flight[110][110];

10 int city[110];

11 int dp[110];

12 int pre[110];

13 int route[110];

14 

15 void putroute(int p)

16 {

17     if(p == 0)

18         return;

19     putroute(pre[p]);

20     printf("%d->", p);

21 }

22 

23 int main(void)

24 {

25     #ifdef LOCAL

26         freopen("1224in.txt", "r", stdin);

27     #endif

28 

29     int N, kase;

30     scanf("%d", &N);

31     for(kase = 1; kase <= N; ++kase)

32     {

33         if(kase > 1)    printf("\n");

34 

35         int n;

36         scanf("%d", &n);

37         int i;

38         for(i = 1; i <= n; ++i)

39             scanf("%d", &city[i]);

40         city[n + 1] = 0;

41         

42         int m, from, to;

43         memset(flight, 0, sizeof(flight));

44         scanf("%d", &m);

45         for(i = 1; i <= m; ++i)

46         {

47             scanf("%d%d", &from, &to);

48             flight[from][to] = true;

49         }

50 

51         memset(dp, 0, sizeof(dp));

52         memset(pre, 0, sizeof(pre));

53         int j;

54         for(i = 2; i <= n + 1; ++i)

55             for(j = 1; j < i; ++j)

56                 if(flight[j][i]

57                     && (dp[j] + city[i]) > dp[i])

58                 {

59                     dp[i] = dp[j] + city[i];

60                     pre[i] = j;

61                 }

62                     

63         

64         printf("CASE %d#\n", kase);

65         printf("points : %d\n", dp[n + 1]);

66 

67         printf("circuit : ");

68         putroute(pre[n + 1]);

69         printf("1\n");

70     }

71 

72     return 0;

73 }

代码君