清北学堂模拟赛day7 石子合并加强版

/*
注意到合并三堆需要枚举两个端点，其实可以开一个数组记录合并两堆的结果，标程好像用了一个神奇的优化
*/
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#define ll long long
#define fo(i,l,r) for(int i = l;i <= r;i++)
#define fd(i,l,r) for(int i = r;i >= l;i--)
using namespace std;
const int maxn = ;
ll read(){
    ll x=,f=;
    char ch=getchar();
    while(!(ch>=''&&ch<='')){if(ch=='-')f=-;ch=getchar();};
    while(ch>=''&&ch<=''){x=x*+(ch-'');ch=getchar();};
    return x*f;
}
int n;
ll dp[maxn][maxn],dp2[maxn][maxn],val[maxn],sum[maxn];
int main(){
    freopen("merge.in","r",stdin);
    freopen("merge.out","w",stdout);
    n = read();
    memset(dp,/,sizeof(dp));
    memset(dp2,/,sizeof(dp2));
    fo(i,,n) val[i] = read();
    fo(i,,n){
        dp[i][i] dp2[i][i] = ;
        sum[i] = sum[i-] + val[i];
    }
    for(int l = ;l <= n;l+=){
        for(int i = ;i + l - <= n;i++){
            int j = i + l - ;
            for(int k1 = i;k1 <= j;k1 += ){
                for(int k2 = k1 + ;k2 <= j;k2 += ){
                    dp[i][j] = min(dp[i][j],dp[i][k1] + dp[k1+][k2] + dp[k2+][j] + sum[j] - sum[i-]);
                }
            }
        }
    }
    cout<<dp[][n];
    return ;
} 

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pr;
const double pi=acos(-);
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define Rep(i,u) for(int i=head[u];i;i=Next[i])
#define clr(a) memset(a,0,sizeof a)
#define pb push_back
#define mp make_pair
#define putk() putchar(' ')
ld eps=1e-;
ll pp=;
ll mo(ll a,ll pp){if(a>= && a<pp)return a;a%=pp;if(a<)a+=pp;return a;}
ll powmod(ll a,ll b,ll pp){ll ans=;for(;b;b>>=,a=mo(a*a,pp))if(b&)ans=mo(ans*a,pp);return ans;}
ll gcd(ll a,ll b){return (!b)?a:gcd(b,a%b);}
ll read(){
    ll ans=;
    char last=' ',ch=getchar();
    while(ch<'' || ch>'')last=ch,ch=getchar();
    while(ch>='' && ch<='')ans=ans*+ch-'',ch=getchar();
    if(last=='-')ans=-ans;
    return ans;
}
void put(ll a){
    if(a<)putchar('-'),a=-a;
    int top=,q[];
    while(a)q[++top]=a%,a/=;
    top=max(top,);
    while(top--)putchar(''+q[top+]);
}
//head
#define INF 1000000000
#define N 410
int n,f1[N][N],f2[N][N],a[N];
int main(){
    freopen("merge.in","r",stdin);
    freopen("merge.out","w",stdout);
    n=read();
    rep(i,,n)
        rep(j,,n)f1[i][j]=f2[i][j]=INF;
    rep(i,,n)a[i]=a[i-]+read();
    rep(i,,n)f2[i][i]=;
    rep(i,,n-)f1[i][i+]=a[i+]-a[i-];
    rep(len,,n)
    rep(i,,n-len+){
        int j=i+len-;
        rep(k,i,j-)f2[i][j]=min(f2[i][j],f1[i][k]+f2[k+][j]+a[j]-a[k]);
        rep(k,i,j-)f1[i][j]=min(f1[i][j],f2[i][k]+f2[k+][j]+a[j]-a[i-]);
    }
    cout<<f2[][n]<<endl;
    return ;
}