HDU2697+DP

Wa的版本。。。

/*
DP
dp[i][j]:前i个取某些个且cost不超过j得到的最大价值
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<set>
#include<math.h>
using namespace std;
typedef long long int64;
//typedef __int64 int64;
typedef pair<int64,int64> PII;
#define MP(a,b) make_pair((a),(b)) 
const int maxn = 105;
const int maxm = 10005;
const int inf = 99999099;
const double pi=acos(-1.0);
const double eps = 1e-8;
struct Node{
    int len;
    bool Choosed;
    int maxVal;
}dp[ maxn ][ maxm ];
void init(){
    for( int i=0;i<maxn;i++ ){
        for( int j=0;j<maxm;j++ ){
            dp[i][j].maxVal = 0;
            dp[i][j].len = 0;
            dp[i][j].Choosed = false;
        }
    }
}
int cost[ maxn ];
int main(){
    int T;
    scanf("%d",&T);
    while( T-- ){
        int n,sum;
        scanf("%d%d",&n,&sum);
        //memset( dp,0,sizeof( dp ) );
        init();
        int SumCost = 0;
        int MinCost = inf;
        for( int i=0;i<n;i++ ){
            scanf("%d",&cost[i]);
            SumCost += cost[ i ];
            MinCost = min( MinCost,cost[i] );
        }
        if( MinCost>sum ) {
            printf("0
");
            continue;
        }
        if( MinCost==sum ) {
            printf("1
");
            continue;
        }
        if( sum>=SumCost ){
            printf("%d
",n*n);
            continue;
        }
        
        int ans = 0;
        for( int i=0;i<n;i++ ){
            for( int j=cost[i];j<=sum;j++ ){
                dp[ i ][ j ].maxVal = 1;
                dp[ i ][ j ].len = 1;
                dp[ i ][ j ].Choosed = true;
                ans = 1;
            }
        }//init of dp
        
        for( int i=1;i<n;i++ ){
            for( int j=0;j<=sum;j++ ){
                if( dp[i-1][j].Choosed==false ){
                    if( j>=cost[i] ){
                        if( dp[i-1][j-cost[i]].maxVal+1>=dp[i-1][j].maxVal ){
                            dp[i][j].maxVal = dp[i-1][j-cost[i]].maxVal+1;
                            dp[i][j].len = 1;
                            dp[i][j].Choosed = true;
                        }
                        else {
                            if( dp[i-1][j].maxVal>dp[i][j].maxVal ){
                                dp[i][j].maxVal = dp[i-1][j].maxVal;
                                dp[i][j].len = 0;
                                dp[i][j].Choosed = false;
                            }
                        }
                    }
                    else {
                        if( dp[i-1][j].maxVal>dp[i][j].maxVal ){
                            dp[i][j].maxVal = dp[i-1][j].maxVal;
                            dp[i][j].len = 0;
                            dp[i][j].Choosed = false;
                        }
                    }
                }
                else{
                    if( j>=cost[i] ){
                        if( dp[i-1][j-cost[i]].maxVal-dp[i-1][j-cost[i]].len*dp[i-1][j-cost[i]].len+(dp[i-1][j-cost[i]].len+1)*(dp[i-1][j-cost[i]].len+1)>=dp[i-1][j].maxVal ){
                            dp[i][j].maxVal = dp[i-1][j-cost[i]].maxVal-dp[i-1][j-cost[i]].len*dp[i-1][j-cost[i]].len+(dp[i-1][j-cost[i]].len+1)*(dp[i-1][j-cost[i]].len+1);
                            dp[i][j].len = dp[i-1][j-cost[i]].len+1;
                            dp[i][j].Choosed = true;
                        }
                        else{
                            if( dp[i-1][j].maxVal>dp[i][j].maxVal ){
                                dp[i][j].maxVal = dp[i-1][j].maxVal;
                                dp[i][j].len = 0;
                                dp[i][j].Choosed = false;
                            }
                        }
                    }
                    else{
                        if( dp[i-1][j].maxVal>dp[i][j].maxVal ){
                            dp[i][j].maxVal = dp[i-1][j].maxVal;
                            dp[i][j].len = 0;
                            dp[i][j].Choosed = false;
                        }
                    }
                }
                ans = max(ans,dp[i][j].maxVal);
                //printf("dp[%d][%d].val = %d, len = %d, choose = %d 
",i,j,dp[i][j].maxVal,dp[i][j].len,dp[i][j].Choosed);
            }
        }
        printf("%d
",ans);
    }
    return 0;
}

Ac代码。。

dp[i][j]:前i个取某些个且cost不超过j得到的最大价值
dp[ i ][ j ] 通过 1 到 i-1 之间的dp[ k ][ ? ] ( 1<=k<i ) 来更新。

如果 dp[ i ][ j ]:取了第i个，则可能是被dp[ k ][ ? ]+ ??更新，反之则为dp[ i-1 ][ j ]

/*
DP
*/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<set>
#include<math.h>
using namespace std;
typedef long long int64;
//typedef __int64 int64;
typedef pair<int64,int64> PII;
#define MP(a,b) make_pair((a),(b)) 
const int maxn = 105;
const int inf = 0x7fffffff;
const double pi=acos(-1.0);
const double eps = 1e-8;
int dp[ maxn ][ maxn ];
int a[ maxn ];
int main(){
	int T;
	scanf("%d",&T);
	while( T-- ){
		int n,sum;
		scanf("%d%d",&n,&sum);
		int MinCost = inf;
		for( int i=1;i<=n;i++ ){
			scanf("%d",&a[i]);
			if( a[i]<MinCost ) MinCost = a[i];
		}
		if( MinCost>sum ){
			printf("0
");
			continue;
		}
		if( MinCost==sum ){
			puts("1");
			continue;
		}
		int ans = 0;
		memset( dp,0,sizeof( dp ) );
		for( int i=1;i<=n;i++ ){
			for( int j=0;j<=sum;j++ ){
				int cnt = 0;
				for( int k=1;k<=i;k++ ){
					cnt += a[ i+1-k ];
					if( cnt>j ) break;
					dp[i][j] = max( dp[i][j],dp[i-k][j-cnt]+k*k );
				}
				dp[i][j] = max( dp[i][j],dp[i-1][j] );
				ans = max( ans,dp[i][j] );
			}
		}
		printf("%d
",ans);
	}
	return 0;
}