HDU2697+DP

dp[i][j]:从前i个中挑出某些且cost不超过j的最大val。

dp[i][j]:应该有1到i-1的dp[k][j-?]来更新！！

 1 /*
 2 DP
 3 dp[i][j]:从前i个中挑出某些且cost不超过j的最大val。
 4 */
 5 #include<stdio.h>
 6 #include<string.h>
 7 #include<stdlib.h>
 8 #include<algorithm>
 9 #include<iostream>
10 #include<queue>
11 #include<map>
12 #include<stack>
13 #include<set>
14 #include<math.h>
15 using namespace std;
16 typedef long long int64;
17 //typedef __int64 int64;
18 typedef pair<int64,int64> PII;
19 #define MP(a,b) make_pair((a),(b)) 
20 const int maxn = 105;
21 const int inf = 0x7fffffff;
22 const double pi=acos(-1.0);
23 const double eps = 1e-8;
24 int dp[ maxn ][ maxn ];
25 int a[ maxn ];
26 int main(){
27     int T;
28     scanf("%d",&T);
29     while( T-- ){
30         int n,sum;
31         scanf("%d%d",&n,&sum);
32         int MinCost = inf;
33         for( int i=1;i<=n;i++ ){
34             scanf("%d",&a[i]);
35             if( a[i]<MinCost ) MinCost = a[i];
36         }
37         if( MinCost>sum ){
38             printf("0\n");
39             continue;
40         }
41         if( MinCost==sum ){
42             puts("1");
43             continue;
44         }
45         int ans = 0;
46         memset( dp,0,sizeof( dp ) );
47         for( int i=1;i<=n;i++ ){
48             for( int j=0;j<=sum;j++ ){
49                 int cnt = 0;
50                 for( int k=1;k<=i;k++ ){
51                     cnt += a[ i+1-k ];
52                     if( cnt>j ) break;
53                     dp[i][j] = max( dp[i][j],dp[i-k][j-cnt]+k*k );
54                 }
55                 dp[i][j] = max( dp[i][j],dp[i-1][j] );
56                 ans = max( ans,dp[i][j] );
57             }
58         }
59         printf("%d\n",ans);
60     }
61     return 0;
62 }
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posted @ 2013-08-13 21:18  xxx0624  阅读(191)  评论(0)  编辑  收藏  举报