bzoj1087

又一道状压dp

查了半天错又被位运算的优先级坑了

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #include<ctime>
 5 #include<cstdlib>
 6 #include<iostream>
 7 #include<algorithm>
 8 #define b(a) (1<<(a))
 9 #define clr(a,x) memset(a,x,sizeof(a))
10 #define rep(i,l,r) for(int i=l;i<r;i++)
11 typedef long long ll;
12 using namespace std;
13 int read()
14 {
15     char c=getchar();
16     int ans=0,f=1;
17     while(!isdigit(c)){
18         if(c=='-') f=-1;
19         c=getchar();
20     }
21     while(isdigit(c)){
22         ans=ans*10+c-'0';
23         c=getchar();
24     }
25     return ans*f;
26 }
27 const int maxn=10,maxk=100;
28 int n,k;
29 ll d[maxn][maxk][b(maxn)];
30 bool legal(int s)
31 {
32     while(s){
33         if(s&1){
34             if(s&b(1)) return 0;
35         }
36         s=s>>1;
37     }
38     return 1;
39 }
40 int cnt(int a)
41 {
42     int sum=0;
43     while(a){
44         if(a&1) sum++;
45         a=a>>1;
46     }
47     return sum;
48 }
49 ll dp(int a,int b,int s)
50 {    
51     if(d[a][b][s]!=-1) return d[a][b][s];
52     if(b==0&&cnt(s==0)) return d[a][b][s]=1;
53     ll sum=0;
54     if(b-cnt(s)>=0){
55         rep(i,0,b(n)){
56             if(legal(i)&&legal(i|s)&&((i&s)==0))
57                 sum+=dp(a-1,b-cnt(s),i);
58         }
59     }    
60     return d[a][b][s]=sum;
61 }
62 int main()
63 {
64     n=read(),k=read();
65     clr(d,-1);
66     rep(i,0,b(n))
67         rep(j,0,k+1)
68             d[1][j][i]=0;
69     rep(i,0,b(n)){
70         if(legal(i)) d[1][cnt(i)][i]=1;
71     } 
72     ll ans=0;
73     rep(i,0,b(n)) 
74     if(legal(i))
75     ans+=dp(n,k,i);
76     printf("%lld
",ans);
77     return 0;
78 }

View Code

1087: [SCOI2005]互不侵犯King

Time Limit: 10 Sec Memory Limit: 162 MB
Submit: 1977 Solved: 1174
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Description

在N×N的棋盘里面放K个国王，使他们互不攻击，共有多少种摆放方案。国王能攻击到它上下左右，以及左上左下右上右下八个方向上附近的各一个格子，共8个格子。

bzoj1087

1087: [SCOI2005]互不侵犯King

Description

Input

Output

Sample Input

Sample Output

HINT

Source