[SDOI2016]征途

题目描述：

雾.

题目分析：

朴素DP O（M*N^2）60

dp[i][p]=min(dp[j][p−1]+(s[j]−s[i])∗(s[j]−s[i]))dp[i][p]=min(dp[j][p−1]+(s[j]−s[i])∗(s[j]−s[i]))

斜率优化 O（M*N）100

dp[k]+(sum[i]−sum[k])2<dp[j]+(sum[i]−sum[j])2dp[k]+(sum[i]−sum[k])2<dp[j]+(sum[i]−sum[j])2

dp[k]−2∗sum[i]∗sum[k]+sum[k]2<dp[j]−2∗sum[i]∗sum[j]+sum[j]2dp[k]−2∗sum[i]∗sum[k]+sum[k]2<dp[j]−2∗sum[i]∗sum[j]+sum[j]2

(dp[k]+sum[k]2)−(dp[j]+sum[j]2)<2∗sum[i]∗sum[k]−2∗sum[i]∗sum[j](dp[k]+sum[k]2)−(dp[j]+sum[j]2)<2∗sum[i]∗sum[k]−2∗sum[i]∗sum[j]

(dp[k]+sum[k]2)−(dp[j]+sum[j]2)<2∗sum[i]∗(sum[k]−sum[j])(dp[k]+sum[k]2)−(dp[j]+sum[j]2)<2∗sum[i]∗(sum[k]−sum[j])

((dp[k]+sum[k]2)−(dp[j]+sum[j]2))/(2∗(sum[k]−sum[j]))<sum[i]((dp[k]+sum[k]2)−(dp[j]+sum[j]2))/(2∗(sum[k]−sum[j]))<sum[i]

题目链接：

Luogu 4072

BZOJ 4518

60朴素DP：

// luogu-judger-enable-o2
#include <cstdio>
#include <iostream>
#include <cstring>
#define sqr(x) x*x
const int maxm=3000+10;
int sum[maxm],dp[maxm][maxm];
int main()
{
//freopen("menci_journey.in","r",stdin);
//freopen("menci_journey.out","w",stdout);
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",sum+i),sum[i]+=sum[i-1];
memset(dp,127/3,sizeof(dp));
dp[0][0]=0;
for(int i=1;i<=n;i++) dp[i][1]=sum[i]*sum[i];
for(int i=1;i<=n;i++)
for(int j=1;j<=std::min(i,m);j++)
for(int k=1;k<i;k++)
dp[i][j]=std::min(dp[i][j],dp[k][j-1]+sqr((sum[i]-sum[k])));

printf("%d\n",m*dp
[m]-sqr(sum
));
return 0;
}

Ac 代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#define il inline
#define ll long long
using namespace std;
const int maxm=3005;
int n,m,a[maxm];
int dl[maxm];
ll dp[maxm][maxm],s[maxm];
il double A(int i, int p) {return (double)dp[i][p] + s[i]*s[i];}
il double slop(int k,int j,int p)
{
return (double)(A(j, p)-A(k, p))/(double)(s[j]-s[k]);
}
int main()
{
//freopen("menci_journey.in","r",stdin);
//freopen("menci_journey.out","w",stdout);
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]),s[i]=s[i-1]+a[i];
dp[0][0]=0;
for(int i=1;i<=n;i++) dp[i][1]=s[i]*s[i];
for(int p=2;p<=m;p++)
{
int l=1,r=0;
for(int i=1;i<=n;i++)
{
while(l<r&&slop(dl[l],dl[l+1],p-1)<2*s[i]) l++;
int j=dl[l];
dp[i][p]=dp[j][p-1]+(s[j]-s[i])*(s[j]-s[i]);
while(l<r&&slop(dl[r],dl[r-1],p-1)>slop(dl[r],i,p-1)) r--;
dl[++r]=i;
}
}
ll ans=m*dp
[m]-s
*s
;
printf("%lld\n",ans);
return 0;
}