Sequence into "APIO 2014"

The meaning of problems

Given a sequence, for \ (K \) operations, each operation will be split sequence.

For each operation, to select any one section divided at any point, and obtained by multiplying the left and right gains.

Seeking maximum returns.

---

Thinking

Substate \ (f [i] [k ] \) representing the forward \ (I \) number of cut \ (K \) times the maximum benefit.

Release equation \ (f [i] [k ] = max (f [j] [k-1] + sum [j] * (sum [i] -sum [j])) \)

Scroll what can be removed one-dimensional.

Suppose decision point \ (x, y \) and \ (X \) than \ (Y \) , it can be simplified to give \ (\ frac {f [x ] -sum [x] ^ 2-f [y ] + SUM [Y] ^ {2} SUM [Y] -sum [X]}> [I SUM] \) . The slope optimization sets click.

// Note that this question will be \ (sum [x] == sum [y] \) case

\(end\)。

Code

#include <bits/stdc++.h>

using namespace std;

namespace StandardIO {

    template<typename T>inline void read (T &x) {
        x=0;T f=1;char c=getchar();
        for (; c<'0'||c>'9'; c=getchar()) if (c=='-') f=-1;
        for (; c>='0'&&c<='9'; c=getchar()) x=x*10+c-'0';
        x*=f;
    }

    template<typename T>inline void write (T x) {
        if (x<0) putchar('-'),x*=-1;
        if (x>=10) write(x/10);
        putchar(x%10+'0');
    }

}

using namespace StandardIO;

namespace Project {
    #define int long long
    
    const int N=100010;
    
    int n,k;
    int head,tail;
    int sum[N],dp[N],dp2[N],queue[N];//,pre[N][205];
    
    inline double slope (int x,int y) {
        if (sum[x]==sum[y]) return -1e18;
        return (double)(dp2[x]-sum[x]*sum[x]-dp2[y]+sum[y]*sum[y])/(double)(sum[y]-sum[x]);
    }

    inline void MAIN () {
        read(n),read(k);
        for (register int i=1; i<=n; ++i) {
            read(sum[i]),sum[i]+=sum[i-1];
        }
        for (register int t=1; t<=k; ++t) {
            head=tail=1,queue[head]=0;
            for (register int i=1; i<=n; ++i) {
                while (head<tail&&slope(queue[head],queue[head+1])<=sum[i]) ++head;
                dp[i]=dp2[queue[head]]+sum[queue[head]]*(sum[i]-sum[queue[head]]);
//              pre[i][t]=queue[head];
                while (head<tail&&slope(queue[tail-1],queue[tail])>=slope(queue[tail],i)) --tail;
                queue[++tail]=i;    
            }
            memcpy(dp2,dp,sizeof(dp));  
        }
        write(dp[n]),putchar('\n');
//      for (register int x=n,i=k; i>=1; --i) x=pre[x][i],write(x),putchar(' ');
    }
    
    #undef int
}

int main () {
//  freopen(".in","r",stdin);
//  freopen(".out","w",stdout);
    Project::MAIN();
}