P1880 [NOI1995]石子合并[区间dp+四边形不等式优化]

P1880 [NOI1995]石子合并

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丢个地址就跑（关于四边形不等式复杂度是n方的证明）

嗯所以这题利用决策的单调性来减少k断点的枚举次数。具体看lyd书。这部分很生疏，但是我还是选择先不管了。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef double db;
 template<typename T>inline char MIN(T&A,T B){return A>B?A=B,:;}
 template<typename T>inline char MAX(T&A,T B){return A<B?A=B,:;}
 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 template<typename T>inline T read(T&x){
     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
 }
 const int N=+,INF=0x3f3f3f3f;
 int p[N][N],Fmin[N][N],Fmax[N][N],sum[N],a[N];
 int n;

 int main(){//freopen("tmp.in","r",stdin);freopen("tmp.out","w",stdout);
     read(n);
     for(register int i=;i<=n;++i)a[i+n]=read(a[i]);
     for(register int i=;i<(n<<);++i)sum[i]=sum[i-]+a[i],p[i][i]=i;
     for(register int i=(n<<)-;i;--i){
         for(register int j=i+;j<(n<<);++j){
             Fmax[i][j]=_max(Fmax[i+][j],Fmax[i][j-])+sum[j]-sum[i-];Fmin[i][j]=INF;
             for(register int k=p[i][j-];k<=p[i+][j];++k)
                 if(MIN(Fmin[i][j],Fmin[i][k]+Fmin[k+][j]+sum[j]-sum[i-]))p[i][j]=k;
             if(!(Fmin[i][j]^INF))Fmin[i][j]=;
         }
     }
     Fmin[][]=INF;
     for(register int i=;i<=n;++i)MIN(Fmin[][],Fmin[i][i+n-]),MAX(Fmax[][],Fmax[i][i+n-]);
     printf("%d\n%d\n",Fmin[][],Fmax[][]);
     return ;
 }