1744: the step jump
time limit: 1000 ms memory limit: 131072 KB
[Title] Description
course side there are N steps in a row, the height of the i-th step is Hi (0 <Hi≤109), the 0-th step, also is the height from the ground zero.
Polo going to put a step into two sets of N Sa, Sb (can be null), a step for the set S = {P1, P2, ... , P | S |}, where P1 <P2 <... < P | S |, he takes
\ (\ sum_ {1} ^ {s} \ left | Hp [i] -Hp [i-1] \ right | \)
of the physical value to complete.
Now he wants to jump twice the minimum value of the overall force required, can you help him?
[Input]
The first number of a row N.
The second row of the N integers Hi.
[] Output
line an integer representing the overall minimum force value.
[Sample] input
. 3
. 1. 1. 3
[Output] Sample
4
[Note]
[data size] and conventions
for 10% of the data N≤20.
For 20% of the data N≤100.
For 50% of the data N≤5000.
To 100% of the data 1≤N≤500000.
【answer】
Dynamic Programming.
The radiation group was found with only about two last step, the definition of f [i] [j] denotes the last two were put is a minimum physical i, j default i> j
When i> when j + 1, i must be placed on the i-1, f [i] [j] = f [i-1] [j] + suan (i-1, i);
当i=j+1时，则f[i][j]可以从f[i-1][k] \(k \in [1,i-1]\)转移。
发现下面的转移是O(n)的，期望得分50。
需要优化。
发现任意一个f[i][j]（i>j+1）必是从f[j+1][j]处转移得到（一直放在第一个集合）。所以只需求出所有f[i-1][j]。
定义sum[i]表示从suan(1,2)+suan(2,3)+……+suan(i-1,i)。
f[i][i-1]=min(f[k][k-1]+sum[i-1]-sum[k]+suan(k-1,i))。
设g[i]=f[i][i-1]，可以将suan里分类讨论分别求出-sum[k]+g[k]+h[k]，和-sum[k]+g[k]-h[k] 按h值压入树状数组内维护前缀最小值即可快速求出从哪儿转移。
代码如下：
#include<bits/stdc++.h> #define int long long using namespace std; const int N=5e5+5; map <int,int> dui; int n,cnt=1,ls[N],a[N]; int g[N],sum[N],sh1[N],sh2[N],ans; inline int read() { char c=getchar(); int x=0; while(!isdigit(c)) c=getchar(); while(isdigit(c)) {x=(x<<3)+(x<<1)+c-'0';c=getchar();} return x; } inline int lowbit(int x) { return x& (-x); } inline void update1(int x,int y) { for(register int i=x;i<=cnt+1;i+=lowbit(i)) sh1[i]=min(sh1[i],y); } inline void update2(int x,int y) { for(register int i=x;i<=cnt+1;i+=lowbit(i)) sh2[i]=min(sh2[i],y); } inline int query1(int x) { int daan=100000000000000; for(register int i=x;i;i-=lowbit(i)) daan=min(daan,sh1[i]); return daan; } inline int query2(int x) { int daan=1000000000000; for(register int i=x;i;i-=lowbit(i)) daan=min(daan,sh2[i]); return daan; } signed main() { n=read(); for(register int i=1;i<=n;i++) { a[i]=read(); sum[i]=sum[i-1]+abs(a[i]-a[i-1]); ls[i]=a[i]; } sort(ls+1,ls+n+1); for(register int i=1;i<=n;i++) { if(ls[i]!=ls[i-1]) dui[ls[i]]=++cnt; } memset(sh1,63,sizeof(sh1)); memset(sh2,63,sizeof(sh2)); g[1]=a[1];ans=sum[n]; update1(1,0); update2(cnt,0); for(register int i=2;i<=n;i++) { int hu1=query1(dui[a[i]]); int hu2=query2(cnt-dui[a[i]]+1); hu1+=a[i];hu2-=a[i]; g[i]=min(hu1,hu2)+sum[i-1]; ans=min(ans,g[i]+sum[n]-sum[i]); update1(dui[a[i-1]],g[i]-sum[i]-a[i-1]); update2(cnt-dui[a[i-1]]+1,g[i]-sum[i]+a[i-1]); } cout<<ans; }