Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Given an alphabet S, and a probability Prob(a) for each , a binary prefix code represents each a in S as a bit string B(a), such that B(a₁) is not a prefix of B(a₂) for any a₁≠a₂ in S.

Huffman’s algorithm constructs a binary prefix code by pairing the two least probable elements of S, a₀ and a₁. a₀ and a₁ are given codes with a common (as yet to be determined) prefix p and differ only in their last bit: B(a₀) = p₀ while B(a₁) = p₁. a₀ and a₁ are removed from S and replaced by a new element b with Prob(b) = Prob(a₀) + Prob(a₁). b is an imaginary element standing for both a₀ and a₁. The Huffman code is computed for this reduced S, and p is set equal to B(b). This reduction of the problem continues until S contains one element a represented by the empty string; that is, when S = {a}, B(a) = ε.

Huffman’s code is optimal in that there is no other prefix code with a shorter average length defined as:

One problem with Huffman codes is that they don’t necessarily preserve any ordering that the elements may have. For example, suppose S = {A, B, C} and Prob(A) = 0.7, Prob(B) = 0.1, Prob(C) = 0.2. A Huffman code for S is B(A) = 1, B(B) = 00, B(C) = 01. The lexicographic ordering of these strings is B(B), B(C), B(A) [i.e. 00,01,1], so the coding does not preserve the original order A, B, C. Therefore, algorithms like binary search might not work as expected on Huffman-coded data.

Given an ordered set S and Prob, you are to compute an ordered prefix code - one whose lexicographic order preserves the order of S.

Input 

Input consists of several data sets. Each set begins with 0<n<100, the number of elements in S. n lines follow; the i-th line gives the probability of a_i, the i-th element of S. Each probability is given as 0.dddd (that is, with exactly four decimal digits). The probabilities sum to 1.0000 exactly. A line containing 0 follows the last data set.

Output 

For each data set, compute an optimal ordered binary prefix code for S. The output should consist of one line giving the average code length, followed by n lines, with the i-th line giving the code for the i-th element of S. If you have solved the problem, these n lines will be in lexicographic order. If there are many optimal solutions, choose any one.

Output an empty line between cases.

Sample Input 

3
0.7000
0.1000
0.2000
3
0.7000
0.2000
0.1000
0

Sample Output 

1.3000
0
10
11
1.3000
0
10
11

---

被題目描述給騙了，寫了霍夫曼提交沒過，後來才發現他要求的是最優二叉樹，
把課本上的代碼敲一敲 O(n^3) 提交就過了。

#include <stdio.h>
int root[105][105];
char bits[105];
void dfs(int l, int r, int dep) {
    if(l > r) {
        bits[dep] = '\0';
        puts(bits);
        return;
    }
    bits[dep] = '0';
    dfs(l, root[l][r]-1, dep+1);
    bits[dep] = '1';
    dfs(root[l][r]+1, r, dep+1);
}
int main() {
    int n;
    while(scanf("%d", &n) == 1 && n) {
        double q[105] = {}, tmp;
        double w[105][105] = {}, e[105][105];
        int l, i, j, r;
        for(i = 0; i < n; i++)
            scanf("%lf", &q[i]);
        n--;
        for(i = 1; i <= n+1; i++)
            e[i][i-1] = w[i][i-1] = q[i-1];
        for(l = 1; l <= n; l++) {
            for(i = 1; i <= n-l+1; i++) {
                j = i+l-1;
                e[i][j] = 32767;
                w[i][j] = w[i][j-1]+q[j];
                for(r = i; r <= j; r++) {
                    tmp = e[i][r-1]+e[r+1][j]+w[i][j];
                    if(tmp < e[i][j]) {
                        e[i][j] = tmp;
                        root[i][j] = r;
                    }
                }
            }
        }
        printf("%.4lf\n", e[1][n]-w[1][n]);
        dfs(1, n, 0);
    }
    return 0;
}