hdu 5691 Sitting in Line (状压dp)

Posted 2020-07-10

 1 #include<bits/stdc++.h>
 2 #define lld I64d
 3 using namespace std;
 4 typedef long long ll;
 5 int cal(int x){
 6     int ret = 0;
 7     while(x){
 8         if(x & 1)  ret ++;
 9         x >>= 1;
10     }
11     return ret;
12 }
13 const int N = 17;
14 const int INF = 0x7f7f7f7f;
15 int dp[1<<16][N];
16 int a[N],p[N],bits[1<<N];
17 /**
18  * dp[i][j] : 表示 状态 i , 最后一个数是 a[j] 的最优值
19  *
20  * a[N] = 0
21  * dp[0][N] = 0 
22  *         dp[i | (1<<k)][k] = max(dp[i | (1 << k)][k],dp[i][j] + a[j] * a[k]);    
23  * */
24 int main(){
25     int n,T,kase = 1;
26     scanf("%d",&T);
27     while(T--){
28         scanf("%d",&n);
29         for(int i = 0 ; i < n ; i ++){
30             scanf("%d%d",&a[i],&p[i]);
31         }
32         for(int i = 0 ; i < (1 << n) ; i ++){
33             for(int j = 0 ; j <= n ; j ++){
34                 dp[i][j] = -INF;
35             }
36         }
37         for(int i = 0 ; i < (1 << n) ; i ++) bits[i] = cal(i);
38         a[n] = dp[0][n] = 0;
39         for(int i = 0 ; i < (1 << n) ; i ++){
40             for(int j = 0 ; j <= n ; j ++){
41                 if(dp[i][j] != -INF){
42                     for(int k = 0 ; k < n ; k ++){
43                         if((i & (1 << k)) == 0 && (p[k] == -1 || p[k] == bits[i]) ){
44                             // 状态转移: 
45                             dp[i | (1<<k)][k] = max(dp[i | (1 << k)][k],dp[i][j] + a[j] * a[k]);
46                         }
47                     }
48                 }
49             }
50         }
　　//  cout << "dp[1][n] :"  << dp[1][n] << endl;    -INF
51         int ans = -INF;
52         for(int i = 0 ; i <= n ; i ++) ans = max(ans,dp[(1<<n)-1][i]);    
53         printf("Case #%d:\n%d\n",kase++,ans);
54     }
55 }

 cout << "dp[1][n] :"  << dp[1][n] << endl;     dp[1][n] = -INF

 

p[k] == -1 表示第a[k]个任意放, p[k] == bits[i] ,表示 a[k]要放在第bits[i]个

 

* a[N] = 0
* dp[0][N] = 0 

dp[1][1] = 0  
　　dp[11][2]
　　dp[101][3]
　　dp[1001][4]
dp[10][2] = 0
　　....
dp[100][3] = 0
dp[1000][4] = 0