BZOJ4426: [Nwerc2015]better productivity Maximum productivity

If a range contains another interval, then the productivity of the two intervals is the same.

Put all the other intervals into the array $b$, the rest into the array $c$, there are more than one time from $b$ to $c$.

Then $c$ all the interval of the left endpoint increment, the right endpoint is also incremented, set $f[i][j]$ for the $c$ in the first $j$ interval divided into $i$ group of the maximum benefit, direct DP can, decision-making has monotonicity.

Then the $p$ assigned to $b$ and $c$, to find out the $b$ and $c$ combination of the maximum benefit can be obtained.

Time Complexity $o (n^2\log N) $.

#include <cstdio> #include <algorithm> #define N 210using namespace Std;int n,m,i,j,flag,cb,cc,s[n],f[n][n] , ans=-2147400000;struct p{int x, y;} A[n],b[n],c[n];bool CMPB (P a,p b) {return a.y-a.x>b.y-b.x;} BOOL CMPC (P a,p b) {return a.x<b.x;}  void dp (int p,int l,int r,int Dl,int dr) {int m= (L+R) >>1,dm=dl,t=ans;    for (int i=min (M-1,DR), i>=dl;i--) {if (c[i+1].y<=c[m].x) break;    int now=f[p-1][i]+c[i+1].y-c[m].x;  if (now>=t) t=now,dm=i;  } f[p][m]=t;  if (L&LT;M) DP (P,L,M-1,DL,DM); if (R&GT;M) DP (P,M+1,R,DM,DR);}  int main () {scanf ("%d%d", &n,&m);  for (i=1;i<=n;i++) scanf ("%d%d", &a[i].x,&a[i].y); for (i=1;i<=n;i++) {for (flag=0,j=1;j<=n;j++) if (a[i].x<=a[j].x&&a[j].y<=a[i].y&& (A[i] . x!=a[j].x| | a[i].y!=a[j].y| |    I<j)) {flag=1;break;}  if (flag) B[++cb]=a[i];else c[++cc]=a[i];  } sort (B+1,B+CB+1,CMPB);  for (i=1;i<=cb;i++) s[i]=s[i-1]+b[i].y-b[i].x;  Sort (C+1,C+CC+1,CMPC);  for (i=1;i<=cc;i++) F[0][i]=ans; for (i=1;i<=m;i++) F[i][0]=ans;  for (i=1;i<=m;i++) DP (I,1,CC,0,CC);  for (i=1;i<=m;i++) if (m-i<=cb&&f[i][cc]>=0) Ans=max (Ans,f[i][cc]+s[m-i]); Return printf ("%d", ans), 0;}

　　

BZOJ4426: [Nwerc2015]better productivity Maximum productivity