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CodeforcesRound#267(Div.2)CGeorgeandJob_html/css_WEB-ITnose

时间:2021-07-01 10:21:17 帮助过:15人阅读

题目大意:从n个数中选出m段不相交的子串,子串的长度均为k,问所有选出来的子串的所有数的和最大为多少。

DP题,DP还是太弱,开始时的dp方程居然写成了O(n^3)...

dp[i][j]: 以num[i]结尾的序列,分成j段的最大和

dp[i][j]=max(dp[k][j-1]+sum[i]-sum[i-m]) 这样的话,其实只要第一重循环是选的段数,第二重循环时数字个数

我又换了种思路

dp[i][j]: 前i个数,分成j段的最大和

dp[i][j]=max(dp[i-1][j],dp[i-m][j-1]+sum[i]-sum[i-m])

思路:二维嘛,所以写出来的dp方程肯定是要么从第一维变化转移过来的,要么从第二维的变化转移过来的


AC代码:

//#pragma comment(linker, "/STACK:102400000,102400000")#include #include #include #include #include #include #include #include #include #include using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i>1;const double EPS = 1e-8;const double pi = acos(-1);const int INF = 100000000;const int MAXN = 5000+100;ll num[MAXN],dp[MAXN][MAXN],pp[MAXN];int n,m,k;ll solve(){    CL(dp,0);    for(int i=m;i<=n;i++)        for(int j=1;j<=k;j++)            dp[i][j]=max(dp[i-m][j-1]+pp[i]-pp[i-m],dp[i-1][j]);    return dp[n][k];}int main(){    //IN("C.txt");    while(~scanf("%d%d%d",&n,&m,&k))    {        pp[0]=0;        for(int i=1;i<=n;i++)        {            scanf("%I64d",&num[i]);            pp[i]=pp[i-1]+num[i];        }        printf("%I64d\n",solve());    }    return 0;}

第一种思路的AC代码(摘自 http://blog.csdn.net/qian99/article/details/39397101):

#include  #include  #include  #include  #include  #include  #include  #include  #include  #include  #include  #define inf 0x3f3f3f3f  #define Inf 0x3FFFFFFFFFFFFFFFLL  #define eps 1e-8  #define pi acos(-1.0)  using namespace std;  typedef long long ll;  const int maxn = 5000 + 5;  int a[maxn];  ll sum[maxn],dp[maxn][maxn],maxv[maxn];  int main()  {  //    freopen("in.txt","r",stdin);  //    freopen("out.txt","w",stdout);      int n,m,k;      scanf("%d%d%d",&n,&m,&k);      for(int i = 1;i <= n;++i)          scanf("%d",&a[i]);      sum[0] = 0;      for(int i = 1;i <= n;++i)          sum[i] = sum[i-1] + a[i];      memset(dp,0xff,sizeof(dp));      memset(maxv,0,sizeof(maxv));      dp[0][0] = 0;      for(int j = 1;j <= k;++j)      {          for(int i = 1;i <= n;++i)          {              if(i - m >= 0)              {                  dp[i][j] = max(dp[i][j],maxv[i-m] + sum[i] - sum[i-m]);              }          }          for(int i = 1;i <= n;++i)              maxv[i] = max(maxv[i-1],dp[i][j]);      }      ll ans = 0;      for(int i = 1;i <= n;++i)          if(dp[i][k] != -1)              ans = max(ans,dp[i][k]);      printf("%I64d\n",ans);      return 0;  }  

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