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  • 【NOIP 2003】 加分二叉树

    【题目链接】

              点击打开链接

    【算法】

               树形DP即可

    【代码】

             

    #include<bits/stdc++.h>
    using namespace std;
    #define MAXN 50
    
    int i,j,k,n,l,r,mid;
    int d[MAXN],f[MAXN][MAXN],root[MAXN][MAXN];
     
    template <typename T> inline void read(T &x) {
            int f = 1; x = 0;
            char c = getchar();
            for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
            for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
            x *= f;
    }
    template <typename T> inline void write(T x) {
        if (x < 0) { putchar('-'); x = -x; }
        if (x > 9) write(x/10);
        putchar(x%10+'0');
    }
    template <typename T> inline void writeln(T x) {
        write(x);
        puts("");
    }
    inline void dfs(int l,int r) {
            if (l <= r) {
                    putchar(' ');
                    write(root[l][r]);
                    dfs(l,root[l][r]-1);
                    dfs(root[l][r]+1,r);    
            }
    }
    
    int main() {
            
            read(n);
            for (i = 0; i <= n; i++) {
                    for (j = 0; j <= n; j++) {
                            f[i][j] = 1;
                    }
            }
            for (i = 1; i <= n; i++) {
                    read(d[i]);
                    f[i][i] = d[i];
                    root[i][i] = i;        
            }
            
            for (k = 2; k <= n; k++) {
                    for (l = 1; l <= n - k + 1; l++) {
                            r = l + k - 1;
                            for (mid = l; mid <= r; mid++) {
                                    if (f[l][mid-1] * f[mid+1][r] + d[mid] > f[l][r]) {
                                            f[l][r] = f[l][mid-1] * f[mid+1][r] + d[mid];
                                            root[l][r] = mid;
                                    }
                            }
                    }    
            }
            writeln(f[1][n]);
            write(root[1][n]);
            dfs(1,root[1][n]-1);
            dfs(root[1][n]+1,n);
            
            return 0;
        
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/9196372.html
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