[刷题]算法竞赛入门经典(第2版) 6-8/UVa806 - Spatial Structures

题意：黑白图像的路径表示法

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代码：（Accepted，0.120s）

//UVa806 - Spatial Structures
//Accepted 0.120s
//#define _XIENAOBAN_
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<cmath>
using namespace std;

int N, T(0);
bool Img[66][66];
vector<string> Sqns;

const unsigned C5_10(const string& five) {//五进制转十进制
    unsigned ans(0), i(1);
    for (auto p(five.rbegin());p != five.rend();++p, i *= 5)
        ans += (*p - 48) * i;
    return ans;
}

const string C10_5(unsigned ten) {//十进制转五进制(倒序)
    string ans;
    do ans += char(ten % 5 + 48);
    while (ten /= 5);
    return ans;
}

bool DFS(int x, int y, string f) {
    int len = N / (int)pow(2, f.length()) / 2;
    if (!len) {
        if (Img[x][y]) {
            Sqns.push_back(f);
            //cerr << "Debug: Sqns(" << x << ", " << y << ")  " << f << endl;//Debug
        }
        return Img[x][y];
    }
    bool NW(DFS(x, y, '1' + f));
    bool NE(DFS(x, y + len, '2' + f));
    bool SW(DFS(x + len, y, '3' + f));
    bool SE(DFS(x + len, y + len, '4' + f));
    bool flag(NW&&NE&&SW&&SE);
    if (flag) {
        Sqns.pop_back(), Sqns.pop_back(), Sqns.pop_back(), Sqns.pop_back();
        Sqns.push_back(f);
    }
    return flag;
}

void Initialize() {
    Sqns.clear();
    for (int i(0);i < N;++i) for (int j(0);j <= N;++j)
        Img[i][j] = false;
}

void SolvePlus() {
    Initialize();

    //Input
    char n;
    for (int i(0);i < N;++i) for (int j(0);j < N;++j)
        cin >> n, Img[i][j] = n - 48;

    //Solve
    DFS(0, 0, "");

    //Output
    if (Sqns.size()) {
        sort(Sqns.begin(), Sqns.end(), [](string& a, string& b)->bool {
            if (a.length() != b.length()) return a.length() < b.length();
            return a < b;
        });
        auto p(Sqns.begin());
        int cnt(0);
        while (p != Sqns.end()) {
            if (cnt == 12) cnt = 0, cout << '\n';
            if (cnt++) cout << ' ' << C5_10(*p);
            else cout << C5_10(*p);
            ++p;
        }
        cout << '\n';
    }
    cout << "Total number of black nodes = " << Sqns.size() << '\n';
}

void SolveMinus() {
    N = -N;
    Initialize();

    //Input
    unsigned n;
    while (cin >> n && n != -1)
        Sqns.push_back(C10_5(n));

    //Solve
    if (Sqns.empty());
    else if (Sqns[0] == "0") {
        for (int i(0);i < N;++i) for (int j(0);j < N;++j)
            Img[i][j] = true;
    }
    else {
        for (const auto& t : Sqns) {
            //cerr <<"Debug1: "<< t << endl;//
            int x(0), y(0), len(N);
            for (const auto& c : t) {
                len /= 2;
                switch (c)
                {
                case '1': break;
                case '2': y += len;break;
                case '3': x += len;break;
                case '4': x += len;y += len;break;
                default:break;
                }
            }
            //cerr << "Debug2: ("<<x << ", " << y <<")  "<<len<< endl;//
            for (int i(0);i < len;++i)
                for (int j(0);j < len;++j)
                    Img[x + i][y + j] = true;
        }
    }

    //Output
    for (int i(0);i < N;++i) {
        for (int j(0);j < N;++j)
            cout << (Img[i][j] ? '*' : '.');
        cout << '\n';
    }
}

int main()
{
#ifdef _XIENAOBAN_
#define gets(T) gets_s(T, 129)
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif

    std::ios::sync_with_stdio(false);
    while (cin >> N && N) {
        if (T) cout << '\n';
        cout << "Image " << ++T << '\n';
        if (N > 0) SolvePlus();
        else SolveMinus();
    }
    return 0;
}

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分析：用的DFS来递归来着，题目倒是不难，直接跟着题意霸王硬上弓就行。但是大神们用10、20ms，我用了120ms。。。刚刚网上看了看好像思路也差不多。啊不管了，其实好久之前写的了，只是忘记发博客里了，我也有点忘了我怎么写的（记得当时写的时候有点昏昏沉沉）。
唉感觉自己越来越懒了，分析也不高兴写了。