結果
問題 | No.74 貯金箱の退屈 |
ユーザー |
|
提出日時 | 2018-01-21 18:10:18 |
言語 | C++17(gcc12) (gcc 12.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 4 ms / 5,000 ms |
コード長 | 6,760 bytes |
コンパイル時間 | 13,052 ms |
コンパイル使用メモリ | 279,052 KB |
最終ジャッジ日時 | 2025-01-05 07:44:10 |
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 30 |
ソースコード
#pragma GCC optimize ("O3")#pragma GCC target ("avx")#include "bits/stdc++.h" // define macro "/D__MAI"using namespace std;typedef long long int ll;#define debugv(v) {printf("L%d %s > ",__LINE__,#v);for(auto e:v){cout<<e<<" ";}cout<<endl;}#define debuga(m,w) {printf("L%d %s > ",__LINE__,#m);for(int x=0;x<(w);x++){cout<<(m)[x]<<" ";}cout<<endl;}#define debugaa(m,h,w) {printf("L%d %s >\n",__LINE__,#m);for(int y=0;y<(h);y++){for(int x=0;x<(w);x++){cout<<(m)[y][x]<<" ";}cout<<endl;}}#define ALL(v) (v).begin(),(v).end()#define repeat(cnt,l) for(auto cnt=0ll;(cnt)<(l);++(cnt))#define rrepeat(cnt,l) for(auto cnt=(l)-1;0<=(cnt);--(cnt))#define iterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);++(cnt))#define MD 1000000007ll#define PI 3.1415926535897932384626433832795template<typename T1, typename T2> ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << "(" << p.first << ":" << p.second << ")"; return o;}template<typename T> T& maxset(T& to, const T& val) { return to = max(to, val); }template<typename T> T& minset(T& to, const T& val) { return to = min(to, val); }void bye(string s, int code = 0) { cout << s << endl; exit(code); }mt19937_64 randdev(8901016);inline ll rand_range(ll l, ll h) {return uniform_int_distribution<ll>(l, h)(randdev);}#if defined(_WIN32) || defined(_WIN64)#define getchar_unlocked _getchar_nolock#define putchar_unlocked _putchar_nolock#elif defined(__GNUC__)#else#define getchar_unlocked getchar#define putchar_unlocked putchar#endifnamespace {#define isvisiblechar(c) (0x21<=(c)&&(c)<=0x7E)class MaiScanner {public:template<typename T> void input_integer(T& var) {var = 0; T sign = 1;int cc = getchar_unlocked();for (; cc<'0' || '9'<cc; cc = getchar_unlocked())if (cc == '-') sign = -1;for (; '0' <= cc && cc <= '9'; cc = getchar_unlocked())var = (var << 3) + (var << 1) + cc - '0';var = var * sign;}inline int c() { return getchar_unlocked(); }inline MaiScanner& operator>>(int& var) { input_integer<int>(var); return *this; }inline MaiScanner& operator>>(long long& var) { input_integer<long long>(var); return *this; }inline MaiScanner& operator>>(string& var) {int cc = getchar_unlocked();for (; !isvisiblechar(cc); cc = getchar_unlocked());for (; isvisiblechar(cc); cc = getchar_unlocked())var.push_back(cc);return *this;}template<typename IT> void in(IT begin, IT end) { for (auto it = begin; it != end; ++it) *this >> *it; }};class MaiPrinter {public:template<typename T>void output_integer(T var) {if (var == 0) { putchar_unlocked('0'); return; }if (var < 0)putchar_unlocked('-'),var = -var;char stack[32]; int stack_p = 0;while (var)stack[stack_p++] = '0' + (var % 10),var /= 10;while (stack_p)putchar_unlocked(stack[--stack_p]);}inline MaiPrinter& operator<<(char c) { putchar_unlocked(c); return *this; }inline MaiPrinter& operator<<(int var) { output_integer<int>(var); return *this; }inline MaiPrinter& operator<<(long long var) { output_integer<long long>(var); return *this; }inline MaiPrinter& operator<<(char* str_p) { while (*str_p) putchar_unlocked(*(str_p++)); return *this; }inline MaiPrinter& operator<<(const string& str) {const char* p = str.c_str();const char* l = p + str.size();while (p < l) putchar_unlocked(*p++);return *this;}template<typename IT> void join(IT begin, IT end, char sep = '\n') { for (auto it = begin; it != end; ++it) *this << *it << sep; }};}MaiScanner scanner;MaiPrinter printer;/*入力をうまく変換すると,コインの状態vector<boolean>と,操作の集合set<pair<int,int>>が与えられる.操作(x,y)を実行すると,x番目のコインとy番目のコインを反転させる.(x==yなら,x番目のコインのみを反転させる)全て表向きにできるか?コインを頂点,操作を辺と置き換えることができる.すると,幾つかの辺を選んで,ww[i]=0の時は奇頂点,ww[i]=1の時は偶頂点となるようにしたい.条件を満たす辺の選び方は存在するか?一筆書きの話に非常によく似ている両端点がww[i]==0,他がww[i]!=0となるようなパスが存在するならば,連続でそのパスを選択することで,パスが通る頂点は条件を満たす.*/class Graph2d {public:typedef int W_T;size_t n;vector<W_T> matrix;Graph2d(size_t size) :n(size), matrix(size*size) {};void resize(size_t s) {n = s; matrix.resize(n*n);}void resize(size_t s, W_T val) {n = s; matrix.resize(n*n, val);}inline W_T& at(int y, int x) { return matrix[y*n + x]; }inline W_T& operator()(int y, int x) { return matrix[y*n + x]; }inline W_T at(int y, int x) const { return matrix[y*n + x]; }inline W_T operator()(int y, int x) const { return matrix[y*n + x]; }inline void connect(int u, int v, W_T dist = 1) {at(u, v) = at(v, u) = dist;}inline void connect_d(int from, int to, W_T dist = 1) { // directedEdge u->vat(from, to) = dist;}};void warshall_floyd(Graph2d& g) {int i, j, k;for (i = 0; i < g.n; i++) {for (j = 0; j < g.n; j++) {for (k = 0; k < g.n; k++) {g(j, k) = min(g(j, k), g(j, i) + g(i, k));}}}}ll m, n, kei;int dd[111];int ww[111];vector<int> flip_from[111]; // iを反転すると,const int inf = 1e9;int main() {scanner >> n;scanner.in(dd, dd + n);scanner.in(ww, ww + n);Graph2d graph(n);fill(ALL(graph.matrix), inf);repeat(i, n) {dd[i] %= n;int x = (i + dd[i]) % n;int y = (i - dd[i] + n) % n;if (x == y) {ww[x] = -1;}else {graph.connect(x, y);}}warshall_floyd(graph);repeat(i, n) {if (ww[i] != 0) continue;iterate(j, 0, n) {if (i == j) continue;if (ww[j] == 1) continue;if (graph(i, j) < inf) {if (ww[j] == 0)ww[j] = 1;ww[i] = 1;break;}}if (ww[i] == 0)bye("No");}bye("Yes");return 0;}