codevs 1154 能量项链(区间dp)

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题目链接：点这里！！！！

题解：

因为是环形，我们就把长度扩大一倍。

然后我们设dp[i][j]表示消除i~j区间剩余a[i]的最大值为多少？

则转移方程：dp[i][j]=max(dp[i][k],dp[k+1][j]+a[i]*a[k+1]*a[j])。

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<sstream>#include<algorithm>#include<vector>#include<bitset>#include<set>#include<queue>#include<stack>#include<map>#include<cstdlib>#include<cmath>#define LL long long#define pb push_back#define pa pair<int,int>#define clr(a,b) memset(a,b,sizeof(a))#define lson lr<<1,l,mid#define rson lr<<1|1,mid+1,r#define bug(x) printf("%d++++++++++++++++++++%d\n",x,x)#define key_value ch[ch[root][1]][0]#pragma comment(linker, "/STACK:102400000000,102400000000")const LL  MOD = 1000000007;const int N = 200+15;const int maxn = 1e5+15;const int letter = 130;const LL INF = 1e7;const double pi=acos(-1.0);const double eps=1e-10;using namespace std;inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}int n,dp[N][N],a[N];int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++) scanf("%d",a+i),a[i+n]=a[i];    for(int len=2;len<=2*n;len++){        for(int i=1;i<=2*n;i++){            int j=i+len-1;            if(j>2*n) break;            for(int k=i;k<j;k++){                dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]+a[i]*a[k+1]*a[j+1]);            }        }    }    int max1=0;    for(int i=1;i<=n;i++) max1=max(max1,dp[i][i+n-1]);    printf("%d\n",max1);    return 0;}

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