【题解】 bzoj1911: [Apio2010]特别行动队 (动态规划+斜率优化)

bzoj1911,懒得复制,戳我戳我

Solution:

  • 线性DP(打牌)
  • \(dp\)方程还是很好想的:\(dp[i]=dp[j-1]+a*(s[i]-s[j-1])^2+b*(s[i]-s[j-1])+c\)
  • 我们假定\(j<k\),且令\(f(j)=dp[j-1]+a*(s[i]-s[j-1])^2+b*(s[i]-s[j-1])+c\)
  • 可以列出式子$$f(j)<f(k)$$
    即(下面这个太长了,自己写写看得清楚些)

\[dp[j-1]+a*(s[i]-s[j-1])^2+b*(s[i]-s[j-1])+c<dp[k-1]+a*(s[i]-s[k-1])^2+b*(s[i]-s[k-1])+c \]

化简可得:

\[\frac{dp[j-1]-dp[k-1]+a*(s[j-1]^2-s[k-1]^2)-b*(s[j-1]-s[k-1])}{2*a*(s[j-1]-s[k-1])}>s[i] \]

  • 然后维护一个下凸包就好了

Code:

//It is coded by Ning_Mew on 5.24
#include<bits/stdc++.h>
#define LL long long
using namespace std;
 
const int maxn=1e6+7;
 
int n;
LL a,b,c,s[maxn],ss=0,tt=1,team[maxn];
LL dp[maxn];
 
double slope(int j,int k){
  return 1.0*(dp[j-1]-dp[k-1]+a*(s[j-1]*s[j-1]-s[k-1]*s[k-1])-b*(s[j-1]-s[k-1]))/(2*a*(s[j-1]-s[k-1])); 
}
LL add(int i,int j){
  LL x=s[j]-s[i-1];return x*x*a+x*b+c;
}
int main(){
  scanf("%d",&n);
  scanf("%lld%lld%lld",&a,&b,&c);
  for(int i=1;i<=n;i++){
    scanf("%lld",&s[i]);
    s[i]+=s[i-1];
  }
  dp[1]=s[1]*s[1]*a+s[1]*b+c;
  team[1]=1;ss=1;tt=2;
  //cout<<"dp:"<<1<<' '<<dp[1]<<endl;
   
  for(int i=2;i<=n;i++){
    while(ss+1<tt&&slope(team[ss],team[ss+1])<s[i])ss++;
    dp[i]=max( dp[team[ss]-1]+add(team[ss],i) , dp[i-1]+add(i,i) );
    //cout<<"dp:"<<i<<' '<<dp[i]<<endl;
    while(ss+1<tt&&slope(team[tt-1],i)<slope(team[tt-2],team[tt-1]))
      tt--;
    team[tt]=i;tt++;  
  }
  printf("%lld\n",dp[n]);
  return 0;
}
posted @ 2018-06-21 18:23  Ning_Mew  阅读(109)  评论(0编辑  收藏  举报