HDU2697+DP

来源：互联网发布：极限矩阵一体机编辑：程序博客网时间：2024/05/14 15:27

Wa的版本。。。

/*DPdp[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\n");            continue;        }        if( MinCost==sum ) {            printf("1\n");            continue;        }        if( sum>=SumCost ){            printf("%d\n",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 \n",i,j,dp[i][j].maxVal,dp[i][j].len,dp[i][j].Choosed);            }        }        printf("%d\n",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\n");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\n",ans);}return 0;}