[状压dp]HDOJ1565 方格取数(1)

中文题~~ 题意略

$n\le 20$ ! 很明显是状压!

 

 1 #include <cstdio>

 2 #include <cstdlib>

 3 #include <cstring>

 4 #include <climits>

 5 #include <cctype>

 6 #include <cmath>

 7 #include <string>

 8 #include <sstream>

 9 #include <iostream>

10 #include <algorithm>

11 #include <iomanip>

12 using namespace std;

13 #include <queue>

14 #include <stack>

15 #include <vector>

16 #include <deque>

17 #include <set>

18 #include <map>

19 typedef long long LL;

20 typedef long double LD;

21 #define pi acos(-1.0)

22 #define lson l, m, rt<<1

23 #define rson m+1, r, rt<<1|1

24 typedef pair<int, int> PI;

25 typedef pair<int, PI> PP;

26 #ifdef _WIN32

27 #define LLD "%I64d"

28 #else

29 #define LLD "%lld"

30 #endif

31 //#pragma comment(linker, "/STACK:1024000000,1024000000")

32 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}

33 //inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}

34 //inline void print(LL x){printf(LLD, x);puts("");}

35 //inline void read(double &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}

36 

37 int mp[25][25];

38 int dp[2][1<<20];   // n格为一个状态

39 int p[1<<20], d;

40 void pre()       //先记录所有的合法状态 即不相邻

41 {

42     d=0;

43     for(int i=0;i<(1<<20);i++)

44         if((i & (i<<1))==0)

45             p[d++]=i;

46 }

47 int main()

48 {

49 #ifndef ONLINE_JUDGE

50     freopen("in.txt", "r", stdin);

51     freopen("out.txt", "w", stdout);

52 #endif

53     pre();

54     int n;

55     while(~scanf("%d", &n))

56     {

57         for(int i=0;i<n;i++)

58             for(int j=0;j<n;j++)

59                 scanf("%d", &mp[i][j]);

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

61         int cur=0;

62         for(int i=0;i<n;i++)

63         {

64             for(int k=0;k<d;k++)

65             {

66                 int sum=0;

67                 if(p[k]>=(1<<n))    // 算是优化吧 没有就tle了

68                         break;

69                 for(int j=0;j<n;j++)

70                     if(p[k] & (1<<j))

71                         sum+=mp[i][j];

72                 for(int j=0;j<d;j++)

73                 {

74                     if(p[j]>=(1<<n))   // 算是优化吧 没有就tle了

75                         break;

76                     if((p[k] & p[j])==0)

77                         dp[cur][p[k]]=max(dp[cur][p[k]], dp[cur^1][p[j]]+sum);

78                 }

79             }

80             cur^=1;

81         }

82         cur^=1;

83         int ans=0;

84         for(int i=0;i<(1<<n);i++)

85             ans=max(ans, dp[cur][i]);

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

87     }

88     return 0;

89 }

HDOJ 1565