[凸包 三分 数形结合] BZOJ 3203 [Sdoi2013]保护出题人

可以发现yi=MAXj<=i{sum[i]−sum[j−1]x[i]+(i−j)∗d}

这个东西是个斜率的形式 (x[i]+i∗d,sum[i])和(sum[j−1],j∗d)

可以发现斜率最大一定在凸包上 维护下凸包 三分最大值

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ld;

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;
}
inline void read(ll &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=100005;

struct P{
ll x,y;
P(ll x=0,ll y=0):x(x),y(y) { }
ld k() { return (ld)y/x; }
friend P operator + (P A,P B) { return P(A.x+B.x,A.y+B.y); }
friend P operator - (P A,P B) { return P(A.x-B.x,A.y-B.y); }
friend ld operator * (P A,P B) { return (ld)A.x*B.y-(ld)B.x*A.y; }
}p
; int pnt;

inline void insert(P np){
while (pnt>1 && (np-p[pnt-1])*(p[pnt]-p[pnt-1])>0) pnt--;
p[++pnt]=np;
}
inline ld query(P q){
int L=1,R=pnt;
while (L+12<R){
int m1=(L+L+R)/3,m2=(L+R+R)/3;
if ((q-p[m1]).k()<(q-p[m2]).k())
L=m1;
else
R=m2;
}
ld ans=0;
for (int i=L;i<=R;i++)
ans=max(ans,(q-p[i]).k());
return ans;
}

int n; ld Ans;
ll d,x
,s
;

int main(){
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(n); read(d);
for (int i=1;i<=n;i++) read(s[i]),read(x[i]),s[i]+=s[i-1];
insert(P(d,s[0]));
for (int i=1;i<=n;i++){
Ans+=query(P(x[i]+d*i,s[i]));
insert(P(d*(i+1),s[i]));
}
printf("%.0lf\n",(double)Ans);
return 0;
}