code2102 石子归并2

划分dp

注意环形，需要把原数复制成两份再进行，详见：

http://www.cnblogs.com/FuTaimeng/p/5427426.html

初始条件：dp[i][i]=0

转移方程：dp[i][j] = max/min{ dp[i][u]+dp[u+1][j]+sum(i,j) }

答案：max/min{ dp[i][i+n-1] }    i=1 to n+1

代码如下：

#include<iostream>
#include<cstring>
#define Size 105
using namespace std;

int n;
int a[Size];
int dp1[Size][Size];//min
int dp2[Size][Size];//max
int sum[Size]; 

inline int s(int l,int r){return sum[r]-sum[l-1];}

int main(){
    memset(dp1,0x3f,sizeof(dp1));
    memset(dp2,-1,sizeof(dp2));
    
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        a[i+n]=a[i];
    }
    for(int i=1;i<=2*n;i++){
        sum[i]=sum[i-1]+a[i];
        dp1[i][i]=dp2[i][i]=0;
    }
    for(int i=1;i<n*2;i++){
        dp1[i][i+1]=dp2[i][i+1]=a[i]+a[i+1];
    }
    
    for(int len=3;len<=n;len++){
        for(int i=1;i+len-1<=n*2;i++){
            int j=i+len-1;
            //dp1[i][j]=min(dp1[i+1][j],dp1[i][j-1])+s(i,j);
            //dp2[i][j]=max(dp2[i+1][j],dp2[i][j-1])+s(i,j);
            for(int u=i;u<j;u++){
                dp1[i][j]=min(dp1[i][j], dp1[i][u]+dp1[u+1][j]+s(i,j));
                dp2[i][j]=max(dp2[i][j], dp2[i][u]+dp2[u+1][j]+s(i,j));
            }
            //cout<<i<<","<<j<<":"<<dp1[i][j]<<"  "<<dp2[i][j]<<endl;
        }
        //cout<<endl;
    }
    
    int ans1=0x3f3f3f3f,ans2=-0x3f3f3f3f;
    for(int i=1;i<=n+1;i++){
        ans1=min(ans1,dp1[i][i+n-1]);
        ans2=max(ans2,dp2[i][i+n-1]);
    }
    
    cout<<ans1<<endl;
    cout<<ans2<<endl;
    
    return 0;
}