[SDOI2016]征途

[SDOI2016]征途

题意

链接

 

题解

DP方程很好推

令dp i,j 表示前I份分到J段

对于一个i,该方程有决策单调性（二次函数）

分治优化即可

#include<bits/stdc++.h>

using namespace std;
 
#define ll long long
#define re register
 
inline ll read()
{
    ll f = 1,x = 0;
    char ch;
    do
    {
        ch = getchar();
        if(ch == '-') f = -1;
    }while(ch<'0'||ch>'9');
    do
    {
        x = (x<<3) + (x<<1) + ch - '0';
        ch = getchar();
    }while(ch>='0'&&ch<='9');
    return f*x;
}

const int MAXN = 3000 + 10;

int n,m;
ll d[MAXN];
ll dp[MAXN][MAXN];
ll ans;

inline void solve(int l,int r,int L,int R,int x)
{
    if(l>r||L>R) return;
    int mid = (l+r)>>1,k = 0;
    dp[x][mid] = 1LL<<60;
    for(int i=L;i<=R&&i<mid;i++)
    {
        ll cur = dp[x-1][i] + (d[mid] - d[i]) * (d[mid] - d[i]);
        if(cur < dp[x][mid]) dp[x][mid] = cur,k = i;
    }
    solve(l,mid-1,L,k,x);solve(mid+1,r,k,R,x);
}

int main()
{
    n = read(),m = read();
    for(re int i=1;i<=n;i++) d[i] = read(),d[i]+=d[i-1];
    for(re int i=1;i<=n;i++) dp[1][i] = d[i] * d[i];
    for(re int i=2;i<=m;i++) solve(1,n,0,n,i);
    cout << dp[m][n] * m - d[n] * d[n] << endl;
}