[DP 决策单调性 || 贪心] UER #7 A.短路

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我是靠谱客的博主冷傲鸵鸟，最近开发中收集的这篇文章主要介绍[DP 决策单调性 || 贪心] UER #7 A.短路，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

先是写了个决策单调性DP

然后知道能贪心就没写了

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

inline char nc(){
  static char buf[100000],*p1=buf,*p2=buf;
  if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
  return *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=100005;

int n,a[N];
ll f[N],sum[N];

struct abcd{
  int x,l,r;
  abcd(int x=0,int l=0,int r=0):x(x),l(l),r(r) { }
}Stack[N];
int pnt;

inline ll Calc(int s,int t){
  if (s>t) return 1LL<<60;
  return f[s]+sum[t]-sum[s]+(ll)(t-s)*a[s];
}

int main(){
  freopen("t.in","r",stdin);
  freopen("t.out","w",stdout);
  read(n); n++;
  for (int i=1;i<=n;i++) read(a[i]);
  reverse(a+1,a+n+1);
  for (int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i];
  Stack[++pnt]=abcd(1,2,n); f[1]=a[1];
  ll ans=(ll)a[1]*(4*n-3);
  for (int i=2;i<=n;i++){
    f[i]=Calc(Stack[pnt].x,i);
    ans=min(ans,f[i]*2+(ll)((n-i)*4-1)*a[i]);
    if (i==n) continue;
    
    while (Stack[pnt].l>i && Calc(Stack[pnt].x,Stack[pnt].l)>=Calc(i,Stack[pnt].l)) pnt--;
    if (Stack[pnt].l<=i && Calc(Stack[pnt].x,i+1)>=Calc(i,i+1)){
      Stack[pnt].r=i;
      Stack[++pnt]=abcd(i,i+1,n);
      continue;
    }
    if (Calc(Stack[pnt].x,Stack[pnt].r)<Calc(i,Stack[pnt].r)) continue;
    int L=max(Stack[pnt].l,i+1)-1,R=Stack[pnt].r,MID;
    while (L+1<R){
      MID=(L+R)>>1;
      if (Calc(Stack[pnt].x,MID)<Calc(i,MID))
	L=MID;
      else
	R=MID;
    }
    Stack[pnt].r=R-1;
    Stack[++pnt]=abcd(i,R,n);
  }
  printf("%lldn",ans);
  return 0;
}