Bzoj 1957 land purchase

This is my first question of slope optimization, which is a souvenir.

First, perform filtering, and then the dynamic planning expression will soon write out f (I) = min (f (I) + B [J + 1] * A [I]).

Then we need to optimize the slope. Obviously, everything here is monotonous, so we only need to maintain the monotonous queue.

The cal function calculates the slope. For details, refer to the Code (refer to others). You need to exercise more in the future.

 1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 using namespace std; 5 long long f[500005]; 6 int a[500005]; 7 int n; 8 long long u[50005],v[50005]; 9 struct node10 {11     long long x,y;12 }p[50005];13 int q[50005];14 int head,tail;15 bool cmp(int x,int y)16 {17     if(u[x]!=u[y])return u[x]<u[y];18     else return v[x]<v[y];19 }20 int m;21 double cal(int x,int y)22 {23     return (double)(f[y]-f[x])/(p[x+1].y-p[y+1].y);24 }25 int main()26 {27     scanf("%d",&n);28     for(int i=1;i<=n;i++)29     {30         a[i]=i;31         scanf("%lld%lld",&u[i],&v[i]);32     }33     sort(a+1,a+n+1,cmp);34     for(int i=1;i<=n;i++)35     {36         while(m && p[m].y<=v[a[i]])m--;37         m++;38         p[m].x=u[a[i]];39         p[m].y=v[a[i]];40     }41     n=m;42     for(int i=1;i<=n;i++)43     {44         while(head<tail && cal(q[head],q[head+1])<p[i].x)head++;45         int t=q[head];46         f[i]=f[t]+p[t+1].y*p[i].x;47         while(head<tail && cal(q[tail-1],q[tail])>cal(q[tail],i))tail--;48         q[++tail]=i;49     }50     printf("%lld\n",f[n]);51     return 0;52 }53     

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Bzoj 1957 land purchase