題目大意

給你一個序列 \(\{c_n\}\)，以及一個正整數 \(M\)，現在要將這個序列分割成連續的若干段，每一段的價值是 \(\left(\sum_{i=1}^{k}c_i\right)^2+M\)，求最小价值。\((n\leq 500000,M\leq 1000)\)

題解

令 \(Sum[n]=\sum_{i=1}^{n}c_i\)，
設 \(dp[i]\) 表示把前 \(i\) 個數分割成若干段的最小价值，那麼有 \(dp[i]=\min\{dp[j]+\left(Sum[i]-Sum[j]\right)^2+M\},j<i\)。
整理後得 \(dp[j]+Sum[j]^2=2Sum[i]Sum[j]+dp[i]-Sum[i]^2-M\)

Code

#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <vector>
using namespace std;

#define RG register int
#define LL long long

template<typename elemType>
inline void Read(elemType &T){
    elemType X=0,w=0; char ch=0;
    while(!isdigit(ch)) {w|=ch=='-';ch=getchar();}
    while(isdigit(ch)) X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
    T=(w?-X:X);
}

LL dp[500005],Sum[500005];
int Q[500005];
int N,head,tail;LL M;

inline LL x(int i){return Sum[i];}
inline LL y(int i){return dp[i]+Sum[i]*Sum[i];}
inline LL k(int i){return Sum[i]<<1;}

inline void maintain(int i,int j){
    dp[i]=dp[j]+(Sum[i]-Sum[j])*(Sum[i]-Sum[j])+M;
}

LL Solve(){
    head=1;tail=0;
    Q[++tail]=0;
    for(RG i=1;i<=N;++i){
        while(tail-head+1>=2 && 
            y(Q[head+1])-y(Q[head])<=k(i)*(x(Q[head+1])-x(Q[head]))) ++head;
        maintain(i,Q[head]);
        while(tail-head+1>=2 && 
            (y(Q[tail])-y(Q[tail-1]))*(x(i)-x(Q[tail-1]))
            >=(y(i)-y(Q[tail-1]))*(x(Q[tail])-x(Q[tail-1]))) --tail;
        Q[++tail]=i;
    }
    return dp[N];
}

int main(){
    while(~scanf("%d%lld",&N,&M)){
        memset(dp,0x3f,sizeof(dp));
        dp[0]=0;
        for(RG i=1;i<=N;++i){
            Read(Sum[i]);
            Sum[i]+=Sum[i-1];
        }
        printf("%lld\n",Solve());
    }
    return 0;
}