結果

問題 No.762 PDCAパス
ユーザー mai
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
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 38
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

/*
*
*/
#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-12
template<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
#endif
namespace {
#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;
// =============================================================================
//
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 0123
constexpr inline int convert(char c) {
return
c == 'P' ? 0 :
c == 'D' ? 1 :
c == 'C' ? 2 :
3;
}
int main() {
ll n, m;
//
string label;
scanner >> n >> m;
scanner >> label;
// PDCA
vector<int> group[4];
group[0].reserve(n);
group[1].reserve(n);
group[2].reserve(n);
group[3].reserve(n);
// P,D,C,A
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]);
//
if (ul + 1 != vl)
continue;
graph.connect(u, v);
}
// DP(Pi)
vector<ll> dp(n);
// P(PP1)
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(PA)
for (int j : group[convert('A')])
ans = (ans + dp[j]) % MD;
//
cout << ans << endl;
return 0;
}
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