Problem 1538 - B - Stones II 贪心+DP

还是给你石头n枚，每一枚石头有两个值a和b，每取一个石头，除了这块石头其余所有的石头的a就都减去这个石头的b，问你取了的石头的a的总和最大可以为多少？

先按B从大到小排序

然后DP:

取的话:dp[i][j]=dp[i-1][j-1]+a[i]-b[i]*(j-1) 注意是j-1

不取的话:dp[i][j]=dp[i-1][j];

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define Maxn 1100
struct Inf
{
    int a,b;
}save[Maxn];
LL dp[Maxn][Maxn];
int n;
LL Max(LL a,LL b)
{
    return a>b?a:b;
}
bool cmp(struct Inf a,struct Inf b)
{
    return a.b>b.b;
}
int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);

   while(scanf("%d",&n)&&n)
   {
       for(int i=;i<=n;i++)
           scanf("%lld%lld",&save[i].a,&save[i].b);
       sort(save+,save+n+,cmp);
       for(int i=;i<=n;i++)
                dp[i][]=;
       for(int i=;i<=n;i++)
       {
           for(int j=;j<=i;j++)
                dp[i][j]=max(dp[i-][j-]+save[i].a-save[i].b*(j-),dp[i-][min(j,i-)]);
               //dp[i][j]=Max(dp[i-1][j],dp[i-1][j+1]+save[i].a-save[i].b*j);
       }
       LL ans=;
       for(int i=;i<=n;i++)
        ans=max(ans,dp[n][i]);
       printf("%lld\n",ans);
   }
   return ;
}

另一种状态方程:

dp[i][j]:表示第i堆后还要选j堆能达到的最大值。

dp[i][j]=max(dp[i-1][j],dp[i-1][j+1]+a[i]-b[i]*j) ; //要么选要么不选

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 1100

struct Inf
{
    int a,b;
}save[Maxn];
LL dp[Maxn][Maxn];
int n;

LL Max(LL a,LL b)
{
    return a>b?a:b;
}

bool cmp(struct Inf a,struct Inf b)
{
    return a.b<b.b;
}

int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);

   while(scanf("%d",&n)&&n)
   {
       for(int i=;i<=n;i++)
           scanf("%lld%lld",&save[i].a,&save[i].b);
       memset(dp,-INF,sizeof(dp)); //无效状态
       //printf("%d\n",dp[1][1]);
       sort(save+,save+n+,cmp);  //顺序 肯定对b从小到大比较优
       for(int i=;i<=n;i++)
            dp[][i]=;

       for(int i=;i<=n;i++)
       {
           for(int j=;j<=n;j++)
               dp[i][j]=Max(dp[i-][j],dp[i-][j+]+save[i].a-save[i].b*j);

       }
       printf("%lld\n",dp[n][]);

   }
   return ;
}