Vijos 1456 最小总代价(状态压缩DP)

题目链接

用二维数组表示，第一维表示状态，第二维表示此状态最后一个专递者。

状态转移为 if((~que1[j])&(1<<u))//开始这个判断想错了，一直RE。。。对位运算不熟啊。。

dp[que1[j]+(1<<u)][u] = min(dp[que1[j]+(1<<u)][u],dp[que1[j]][k]+p[k+1][u+1]);

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cmath>

 4 #include <string>

 5 using namespace std;

 6 #define INF 0x3f3f3f3f

 7 int p[20][20];

 8 int dp[500000][17];

 9 int que1[500000],que2[500000];

10 int o[500000];

11 int main()

12 {

13     int n,i,j,k,u,num,t;

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

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

16     {

17         for(j = 1; j <= n; j ++)

18             scanf("%d",&p[i][j]);

19     }

20     for(i = 0; i < 1<<n; i ++)

21     {

22         for(j = 0; j < n; j ++)

23         {

24             dp[i][j] = INF;

25         }

26     }

27     num = 0;

28     for(i = 0; i < n;i ++)

29     {

30         dp[1<<i][i] = 0;

31         que1[num++] = 1<<i;

32     }

33     for(i = 2;i <= n;i ++)

34     {

35         t = 0;

36         for(j = 0;j < num;j ++)

37         {

38             for(k = 0;k < n;k ++)

39             {

40                 for(u = 0;u < n;u ++)

41                 {

42                     if(k != u)

43                     {

44                         if((~que1[j])&(1<<u))

45                         {

46                             dp[que1[j]+(1<<u)][u] = min(dp[que1[j]+(1<<u)][u],dp[que1[j]][k]+p[k+1][u+1]);

47                             if(!o[que1[j]+(1<<u)])

48                             {

49                                 o[que1[j]+(1<<u)] = 1;

50                                 que2[t++] = que1[j]+(1<<u);

51                             }

52                         }

53                     }

54                 }

55             }

56         }

57         num = t;

58         for(j = 0;j < num;j ++)

59         {

60             que1[j] = que2[j];

61             que2[j] = 0;

62         }

63     }

64     int ans = INF;

65     for(i = 0;i < n;i ++)

66     ans = min(dp[(1<<n)-1][i],ans);

67     printf("%d\n",ans);

68     return 0;

69 }

70 /*

71 16

72 -1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

73 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

74 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1 1

75 1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 1

76 1 1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1

77 1 1 1 1 1 -1 1 1 1 1 1 1 1 1 1 1

78 1 1 1 1 1 1 -1 1 1 1 1 1 1 1 1 1

79 1 1 1 1 1 1 1 -1 1 1 1 1 1 1 1 1

80 1 1 1 1 1 1 1 1 -1 1 1 1 1 1 1 1

81 1 1 1 1 1 1 1 1 1 -1 1 1 1 1 1 1

82 1 1 1 1 1 1 1 1 1 1 -1 1 1 1 1 1

83 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 1 1

84 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 1

85 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1

86 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -1 1

87 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 -1

88 */