[BZOJ3963][WF2011][CDQ分治][斜率优化][DP]MachineWorks

被精度卡死……

令d[i]为第i台机器可以购买的日期，f[i]表示第d[i]天的最大收入，

则f[i]=max{f[j]-p[j]+r[j]+(d[i]-d[j]-1)*g[j]}

令A[i]=f[i]-p[i]+r[i]-(d[i]+1)*g[i]

f[i]=A[j]+d[i]g[j]，则A[j]=f[i]-d[i]g[j]

那么就用CDQ分治维护一下凸包就行了

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#define N 400010
#define eps 1e-12

using namespace std;

typedef long long ll;

ll n,c,d,t,q
;
ll f
;
struct stp{
ll d,p,r,g,w,b;
}A
,ia
;
struct Point{
ll x,y;
int ct;
Point(){x=y=ct=0;}
Point(ll a,ll b):x(a),y(b){}
friend bool operator <(Point a,Point b){
return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
friend Point operator -(Point a,Point b){
return Point(a.x-b.x,a.y-b.y);
}
}p
,ib
;

inline double cross(Point a,Point b){
return 1.0*a.x*b.y-1.0*b.x*a.y;
}

inline int dcmp(double x){
return fabs(x)<eps?0:(x<0?-1:1);
}

inline bool cmp(stp a,stp b){
return a.d<b.d;
}

inline ll cal(int x,int y){
return A[x].d*ib[y].x+ib[y].y;
}

inline double getk(int a,int b){
return (double)(ib[a].y-ib[b].y)/(double)(ib[a].x-ib[b].x);
}

void solve(int l,int r){
if(l==r){
f[l]=max(f[l],f[l-1]);
return;
}
int mid=l+r>>1,t1=0;
solve(l,mid);
int head=1,tail=0;
for(int i=l;i<=mid;i++) if(f[i]>=A[i].p) ib[++t1]=Point(A[i].g,f[i]-A[i].p+A[i].r-1ll*(A[i].d+1)*A[i].g);
sort(ib+1,ib+t1+1);
for(int i=1;i<=t1;i++){
while(head<tail&&dcmp(cross(ib[q[tail]]-ib[q[tail-1]],ib[i]-ib[q[tail-1]]))>=0) tail--;
q[++tail]=i;
}
for(int i=mid+1;i<=r;i++){
while(head<tail&&cal(i,q[head])<=cal(i,q[head+1])) head++;
if(head<=tail)f[i]=max(f[i],cal(i,q[head]));
}
solve(mid+1,r);
}

int main(){
while(1){
memset(f,0,sizeof(f));
scanf("%lld%lld%lld",&n,&f[0],&d);
if(!n&&!f[0]&&!d) break;
for(int i=1;i<=n;i++){
scanf("%lld%lld%lld%lld",&A[i].d,&A[i].p,&A[i].r,&A[i].g);
}
A[++n].b=n;A
.d=d+1;
sort(A+1,A+1+n,cmp);
solve(1,n);
printf("Case %lld: %lld\n",++t,f
);
}
}