[区间DP] Codeforces 392E Round #230 (Div. 1) E. Deleting Substrings

好水的E题

f[i][j]表示删完[i,j]区间的最大收益
g[i][j] 表示把[i,j]删成a[i], a[i]+1, a[i]+2 …a[j]的最大收益
h[i][j]类似g[i][j]，递减

f[i][j] <- g[i][k]+h[k][j]+v[2*a[k]-a[i]-a[j]+1]

#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=405;int n,a[N],v[N];int f[N][N],g[N][N],h[N][N];int ans[N];int main(){  freopen("haha.in","r",stdin);  freopen("haha.out","w",stdout);  read(n);  for (int i=1;i<=n;i++) read(v[i]); for (int i=1;i<=n;i++) read(a[i]);  for (int l=1;l<=n;l++)    for (int i=1;i+l-1<=n;i++){      int j=i+l-1;      if (i==j){    g[i][j]=0,h[i][j]=0;    f[i][j]=v[1];    continue;      }      g[i][j]=h[i][j]=-1<<28;      for (int k=i+1;k<=j;k++)    if (a[k]==a[i]+1)      g[i][j]=max(g[i][j],f[i+1][k-1]+g[k][j]);    else if (a[k]==a[i]-1)      h[i][j]=max(h[i][j],f[i+1][k-1]+h[k][j]);      f[i][j]=-1<<28;      for (int k=i;k<j;k++)    f[i][j]=max(f[i][j],f[i][k]+f[k+1][j]);      for (int k=i;k<=j;k++)    if (g[i][k]>-1<<20 && h[k][j]>-1<<20)      f[i][j]=max(f[i][j],g[i][k]+h[k][j]+v[2*a[k]-a[i]-a[j]+1]);    }  for (int i=1;i<=n;i++){    ans[i]=ans[i-1];    for (int j=1;j<=i;j++)      ans[i]=max(ans[i],ans[j-1]+f[j][i]);  }  printf("%d\n",ans[n]);  return 0;}