[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。