結果
問題 | No.762 PDCAパス |
ユーザー |
|
提出日時 | 2017-11-05 13:48:55 |
言語 | C++17(gcc12) (gcc 12.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 61 ms / 2,000 ms |
コード長 | 5,227 bytes |
コンパイル時間 | 14,046 ms |
コンパイル使用メモリ | 281,512 KB |
最終ジャッジ日時 | 2025-01-05 03:48:50 |
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 38 |
ソースコード
/** 解説用*/#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 xprintf(fmt,...) fprintf(stderr,fmt,__VA_ARGS__)#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 iterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);++(cnt))#define MD 1000000007ll#define PI 3.1415926535897932384626433832795#define EPS 1e-12template<typename T1, typename T2> ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << "(" << p.first << ":" << p.second << ")"; return o;}template<typename iterator> inline size_t argmin(iterator begin, iterator end) { return distance(begin, min_element(begin, end)); }template<typename iterator> inline size_t argmax(iterator begin, iterator end) { return distance(begin, max_element(begin, end)); }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(0); }mt19937_64 randdev(8901016);inline ll rand_range(ll l, ll h) {return uniform_int_distribution<ll>(l, h)(randdev);}#ifdef __MAI#define getchar_unlocked getchar#define putchar_unlocked putchar#endif#ifdef __VSCC#define getchar_unlocked _getchar_nolock#define putchar_unlocked _putchar_nolock#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;}};}MaiScanner scanner;// =============================================================================// 隣接頂点を保持する有向グラフclass DGraph {public:size_t n;vector<vector<int>> vertex_to;vector<vector<int>> vertex_from;DGraph(size_t n) :n(n), vertex_to(n), vertex_from(n) {}void connect(int from, int to) {vertex_to[from].emplace_back(to);vertex_from[to].emplace_back(from);}void resize(size_t _n) {n = _n;vertex_to.resize(_n);vertex_from.resize(_n);}};// PDCA を 0123 に置き換えるconstexpr inline int convert(char c) {returnc == 'P' ? 0 :c == 'D' ? 1 :c == 'C' ? 2 :3;}int main() {ll n, m;// 頂点に書き込まれた文字string label;scanner >> n >> m;scanner >> label;// Pと書き込まれた頂点集合,D,C,Avector<int> group[4];group[0].reserve(n);group[1].reserve(n);group[2].reserve(n);group[3].reserve(n);// 各頂点をP,D,C,Aグループに分配repeat(i, n) {group[convert(label[i])].push_back(i);}DGraph graph(n);repeat(i, m) {int u, v;scanner >> u >> v;--u; --v;if (label[u] < label[v])swap(u, v);int ul = convert(label[u]);int vl = convert(label[v]);// 題意とは関係の無い辺を除去if (ul + 1 != vl)continue;graph.connect(u, v);}// DP(あるPに属する頂点から頂点iまで移動する経路は何通りあるか)vector<ll> dp(n);// P(PからPまで移動する経路は1通り)for (int j : group[convert('P')])dp[j] = 1;// (P->D), (D->C), (C->A)repeat(i, 3) {for (int i : group[i])for (int j : graph.vertex_to[i])dp[j] = (dp[j] + dp[i]) % MD;}ll ans = 0;// A(あるPに属する頂点からAまで移動する経路のパターン数の総和が答え)for (int j : group[convert('A')])ans = (ans + dp[j]) % MD;// こたえcout << ans << endl;return 0;}