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  • Codeforces 1025D(区间dp)

      容易想到设f[i][j][k]为i~j区间以k为根是否能构成bst。这样是O(n4)的。考虑将状态改为f[i][j][0/1]表示i~j区间以i-1/j+1为根能否构成bst。显然如果是i-1作为根的话i~j区间都在它的右子树,所以转移时枚举右子树的根并判断是否合法,j+1类似。

    #include<iostream> 
    #include<cstdio>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int read()
    {
        int x=0,f=1;char c=getchar();
        while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
        while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
        return x*f;
    }
    #define N 710
    int n,a[N];
    bool flag[N][N],f[N][N][2];
    int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
    int main()
    {
    #ifndef ONLINE_JUDGE
        freopen("d.in","r",stdin);
        freopen("d.out","w",stdout);
        const char LL[]="%I64d
    ";
    #else
        const char LL[]="%lld
    ";
    #endif
        n=read();
        for (int i=1;i<=n;i++) a[i]=read();
        for (int i=1;i<=n;i++)
            for (int j=1;j<=n;j++)
            if (gcd(a[i],a[j])>1) flag[i][j]=1;
        for (int i=1;i<=n+1;i++) f[i][i-1][0]=f[i][i-1][1]=1;
        for (int k=1;k<=n;k++)
            for (int i=1;i<=n-k+1;i++)
            {
                int j=i+k-1;
                for (int d=i;d<=j;d++)
                if (f[i][d-1][1]&&f[d+1][j][0])
                {
                    if (flag[i-1][d]) f[i][j][0]=1;
                    if (flag[j+1][d]) f[i][j][1]=1;
                }
            }
        for (int i=1;i<=n;i++) if (f[1][i-1][1]&&f[i+1][n][0]) {cout<<"Yes";return 0;}
        cout<<"No";
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Gloid/p/9780193.html
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