Sicily 1001. Alphacode

最新推荐文章于 2019-06-26 23:26:54 发布

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本文探讨了一种特定的消息编码方式，并提出了解决方案来计算不同解码的可能性数量。通过使用动态规划的方法，有效地解决了问题，避免了冗余计算。

题目描述
Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Alice and Bob need to send secret messages to each other and are discussing ways to encode their messages: Alice: “Let’s just use a very simple code: We’ll assign A' the code word 1,B’ will be 2, and so on down to Z' being assigned 26." Bob: "That's a stupid code, Alice. Suppose I send you the wordBEAN’ encoded as 25114. You could decode that in many different ways!” Alice: “Sure you could, but what words would you get? Other than BEAN', you'd getBEAAD’, YAAD',YAN’, YKD' andBEKD’. I think you would be able to figure out the correct decoding. And why would you send me the word `BEAN’ anyway?” Bob: “OK, maybe that’s a bad example, but I bet you that if you got a string of length 500 there would be tons of different decodings and with that many you would find at least two different ones that would make sense.” Alice: “How many different decodings?” Bob: “Jillions!” For some reason, Alice is still unconvinced by Bob’s argument, so she requires a program that will determine how many decodings there can be for a given string using her code.

Input

Input will consist of multiple input sets. Each set will consist of a single line of digits representing a valid encryption (for example, no line will begin with a 0). There will be no spaces between the digits. An input line of `0’ will terminate the input and should not be processed

Output

For each input set, output the number of possible decodings for the input string. All answers will be within the range of a long variable.

Sample Input

25114
1111111111
3333333333
0
Sample Output

6
89

解题思路：
先不管三七二十一，花几分钟直接暴力尝试下，暴力的思想也是很简单，将这个字符串当做一棵树用dfs来遍历即可：

#include <iostream>
#include <string>
using namespace std;

long long res = 0;
string str;

bool canCombineWithNext(size_t cursor)
{
    return str[cursor] == '1' || (str[cursor]=='2' && str[cursor+1]<='6');
}

void dfs(size_t cursor)
{
    if(cursor >= str.size()-1){
        res++;
        return;
    }

    if(canCombineWithNext(cursor)){
        if(str[cursor+1] != '0')
            dfs(cursor+1);
        dfs(cursor+2);
    }
    else{
        dfs(cursor+1);
    }
}

int main()
{
    while(cin >> str){
        if(str[0] == '0')
            break;
        dfs(0);
        cout << res << endl;
        res = 0;
    }
    return 0;
}

结果很明显是time limit error，这主要是由于题目中提到结果值满足long长度，也就是该算法的复杂度达到10^18，但仔细分析可以发现，这其中有大量冗余计算，比如对于25114来说，当确定以2来遍历的时候或者以25开头来遍历的时候，后面有很大部分出现了冗余遍历（也就是都遍历了114这部分），基于此，不难想到可以用动态规划保留前面的遍历结果供后面计算。
具体代码如下，其中dp[i]数组表示从位置0到位置i可以有多少情况。时间复杂度为O(n)，n为输入的字符串的长度。

#include <iostream>
#include <string>
#include <vector>
using namespace std;

string str;
vector<long long> dp;

bool canCombineWithPre(size_t cursor)
{
    return (str[cursor-1] == '1' 
            || (str[cursor-1]=='2' && str[cursor]<='6'));
}

void initalDp()
{
    dp.clear();
    dp.push_back(1);
    if(str.size() > 1 
            && str[1] != '0' && canCombineWithPre(1)){
        dp.push_back(2);
    }
    else{
        dp.push_back(1);
    }
}

void calDp(const size_t cursor)
{
    if(cursor > str.size()-1)
        return;

    long long currentDp = 0;        // 当前游标对应的dp值
    if(canCombineWithPre(cursor)){
        if(str[cursor] == '0')
            currentDp = dp[cursor-2];
        else
            currentDp = dp[cursor-1] + dp[cursor-2];
    }
    else{
        currentDp = dp[cursor-1];
    }
    dp.push_back(currentDp);

    calDp(cursor+1);
}

int main()
{
    while(cin >> str){
        if(str[0] == '0')
            break;
        initalDp();
        calDp(2);
        cout << dp[str.size()-1] << endl;
    }
    return 0;
}