$Noip2018/Luogu5019/Luogu1969$ 铺设道路

$Luogu$

去年$Noip$的时候我并没有做过原题,然后考场上也没有想出正解,就写了个优化了一点的暴力:树状数组+差分,然后就$A$了$ovo$.

$Sol$

只要$O(N)$扫一遍,只要当前值比前一个值大,那么答案就累计这两个值的差的绝对值.$over.$

$Code$

#include<iostream>
#include<cstdio>
#define rg register
#define ll long long 
using namespace std;
int read()
{
  int x=0,y=1;char c;
  c=getchar();
  while(c<'0'||c>'9') {if(c=='-') y=-1;c=getchar();}
  while(c>='0'&&c<='9') {x=(x<<1)+(x<<3)+c-'0';c=getchar();}
  return x*y;
}
int n;
ll ans;
int a[100010],b[100010];
int lowbit(int x)
{
  return x&(-x);
}
void add(int x,int k)
{
  for(rg int i=x;i<=n;i+=lowbit(i)) b[i]+=k;
}
ll sum(int x)
{
  ll sum1=0;
  for(rg int i=x;i>=1;i-=lowbit(i))
    sum1+=b[i];
  return sum1;
}
int minn,l,r;
int main()
{
  freopen("road.in","r",stdin);
  freopen("road.out","w",stdout);
  n=read();minn=210000000;
  for(rg int i=1;i<=n;i++)
    {
      a[i]=read();
      minn=min(a[i],minn);
      add(i,a[i]-a[i-1]);
    }
  ans+=minn;
  add(1,-minn);
  if(a[1]!=0) l=0;
  for(rg int i=1;i<=n;i++)
    {
      int s=sum(i);
      if(s)
    {
      if(l==0) {l=i;r=i;minn=s;}
      else {minn=min(minn,s);r=i;}
      if(i!=n) continue ;
    }
      else if(l==0) continue;
      ans+=minn;
      add(l,-minn);
      if(r<n) add(r+1,minn);
      i=l-1;l=0;
    }
  printf("%lld",ans);
  return 0;
}

#include<iostream>
#include<cstdio>
#define il inline
#define Rg register
#define go(i,a,b) for(Rg int i=a;i<=b;++i)
#define ll long long
using namespace std;
il int read()
{
    Rg int x=0,y=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    return x*y;
}
int n,x,y;ll as;
int main()
{
    n=read();
    go(i,1,n){y=read();if(y>x)as+=y-x;x=y;}
    printf("%lld
",as);
    return 0;
}