結果
| 問題 | No.422 文字列変更 (Hard) |
| ユーザー |
 heno239
|
| 提出日時 | 2022-10-12 13:23:26 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 40 ms / 3,000 ms |
| コード長 | 6,896 bytes |
| コンパイル時間 | 2,098 ms |
| コンパイル使用メモリ | 156,432 KB |
| 最終ジャッジ日時 | 2025-02-08 02:00:49 |
|
ジャッジサーバーID （参考情報） |
judge1 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 1 |
| other | AC * 16 |
ソースコード
#pragma GCC optimize("O3")#pragma GCC optimize("unroll-loops")#include<iostream>#include<string>#include<cstdio>#include<vector>#include<cmath>#include<algorithm>#include<functional>#include<iomanip>#include<queue>#include<ciso646>#include<random>#include<map>#include<set>#include<bitset>#include<stack>#include<unordered_map>#include<unordered_set>#include<utility>#include<cassert>#include<complex>#include<numeric>#include<array>#include<chrono>using namespace std;//#define int long longtypedef long long ll;typedef unsigned long long ul;typedef unsigned int ui;constexpr ll mod = 998244353;//constexpr ll mod = 1000000007;const ll INF = mod * mod;typedef pair<int, int>P;#define rep(i,n) for(int i=0;i<n;i++)#define per(i,n) for(int i=n-1;i>=0;i--)#define Rep(i,sta,n) for(int i=sta;i<n;i++)#define rep1(i,n) for(int i=1;i<=n;i++)#define per1(i,n) for(int i=n;i>=1;i--)#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)#define all(v) (v).begin(),(v).end()typedef pair<ll, ll> LP;template<typename T>void chmin(T& a, T b) {a = min(a, b);}template<typename T>void chmax(T& a, T b) {a = max(a, b);}template<typename T>void cinarray(vector<T>& v) {rep(i, v.size())cin >> v[i];}template<typename T>void coutarray(vector<T>& v) {rep(i, v.size()) {if (i > 0)cout << " "; cout << v[i];}cout << "\n";}ll mod_pow(ll x, ll n, ll m = mod) {if (n < 0) {ll res = mod_pow(x, -n, m);return mod_pow(res, m - 2, m);}if (abs(x) >= m)x %= m;if (x < 0)x += m;//if (x == 0)return 0;ll res = 1;while (n) {if (n & 1)res = res * x % m;x = x * x % m; n >>= 1;}return res;}//mod should be <2^31struct modint {int n;modint() :n(0) { ; }modint(ll m) {if (m < 0 || mod <= m) {m %= mod; if (m < 0)m += mod;}n = m;}operator int() { return n; }};bool operator==(modint a, modint b) { return a.n == b.n; }bool operator<(modint a, modint b) { return a.n < b.n; }modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= (int)mod; return a; }modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += (int)mod; return a; }modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }modint operator+(modint a, modint b) { return a += b; }modint operator-(modint a, modint b) { return a -= b; }modint operator*(modint a, modint b) { return a *= b; }modint operator^(modint a, ll n) {if (n == 0)return modint(1);modint res = (a * a) ^ (n / 2);if (n % 2)res = res * a;return res;}ll inv(ll a, ll p) {return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);}modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }modint operator/=(modint& a, modint b) { a = a / b; return a; }const int max_n = 1 << 21;modint fact[max_n], factinv[max_n];void init_f() {fact[0] = modint(1);for (int i = 0; i < max_n - 1; i++) {fact[i + 1] = fact[i] * modint(i + 1);}factinv[max_n - 1] = modint(1) / fact[max_n - 1];for (int i = max_n - 2; i >= 0; i--) {factinv[i] = factinv[i + 1] * modint(i + 1);}}modint comb(int a, int b) {if (a < 0 || b < 0 || a < b)return 0;return fact[a] * factinv[b] * factinv[a - b];}modint combP(int a, int b) {if (a < 0 || b < 0 || a < b)return 0;return fact[a] * factinv[a - b];}ll gcd(ll a, ll b) {a = abs(a); b = abs(b);if (a < b)swap(a, b);while (b) {ll r = a % b; a = b; b = r;}return a;}using ld = long double;//typedef long double ld;typedef pair<ld, ld> LDP;const ld eps = 1e-10;const ld pi = acosl(-1.0);template<typename T>void addv(vector<T>& v, int loc, T val) {if (loc >= v.size())v.resize(loc + 1, 0);v[loc] += val;}/*const int mn = 2000005;bool isp[mn];vector<int> ps;void init() {fill(isp + 2, isp + mn, true);for (int i = 2; i < mn; i++) {if (!isp[i])continue;ps.push_back(i);for (int j = 2 * i; j < mn; j += i) {isp[j] = false;}}}*///[,val)template<typename T>auto prev_itr(set<T>& st, T val) {auto res = st.lower_bound(val);if (res == st.begin())return st.end();res--; return res;}//[val,)template<typename T>auto next_itr(set<T>& st, T val) {auto res = st.lower_bound(val);return res;}using mP = pair<modint, modint>;mP operator+(mP a, mP b) {return { a.first + b.first,a.second + b.second };}mP operator+=(mP& a, mP b) {a = a + b; return a;}mP operator-(mP a, mP b) {return { a.first - b.first,a.second - b.second };}mP operator-=(mP& a, mP b) {a = a - b; return a;}LP operator+(LP a, LP b) {return { a.first + b.first,a.second + b.second };}LP operator+=(LP& a, LP b) {a = a + b; return a;}LP operator-(LP a, LP b) {return { a.first - b.first,a.second - b.second };}LP operator-=(LP& a, LP b) {a = a - b; return a;}mt19937 mt(time(0));const string drul = "DRUL";string senw = "SENW";//DRUL,or SENWint dx[4] = { 1,0,-1,0 };int dy[4] = { 0,1,0,-1 };//-----------------------------------------int dp[1205][1205];P pre[1205][1205];P dps[1205][1205];P dpt[1205][1205];void solve() {int n, m; cin >> n >> m;string s, t; cin >> s >> t;rep(i, n + 1)rep(j, m + 1) {dp[i][j] = mod;}dp[0][0] = 0;dps[0][0] = { 9,0 };dpt[0][0] = { 9,0 };rep(i, n + 1)rep(j, m + 1) {if (i == 0 && j == 0)continue;if (i > 0 && j > 0) {int cost = 0;if (s[i - 1] != t[j - 1])cost = 5;if (dp[i - 1][j - 1] + cost < dp[i][j]) {dp[i][j] = dp[i - 1][j - 1] + cost;pre[i][j] = { i - 1,j - 1 };}}if (i > 0) {if (dps[i - 1][j].first < dp[i][j]) {dp[i][j] = dps[i - 1][j].first;pre[i][j] = { dps[i - 1][j].second,j };}}if (j > 0) {if (dpt[i][j - 1].first < dp[i][j]) {dp[i][j] = dpt[i][j - 1].first;pre[i][j] = { i,dpt[i][j - 1].second};}}dps[i][j] = { dp[i][j]+9,i };if (i > 0) {if (dps[i - 1][j].first + 2 < dps[i][j].first) {dps[i][j] = dps[i - 1][j];dps[i][j].first += 2;}}dpt[i][j] = { dp[i][j]+9,j };if (j > 0) {if (dpt[i][j - 1].first + 2 < dpt[i][j].first) {dpt[i][j] = dpt[i][j - 1];dpt[i][j].first += 2;}}}string anss, anst;int ci = n, cj = m;while (ci > 0 || cj > 0) {P p = pre[ci][cj];int ni = p.first, nj = p.second;if (ni + 1 == ci && nj + 1 == cj) {anss.push_back(s[ni]);anst.push_back(t[nj]);}else if (ni < ci) {for (int j = ci - 1; j >= ni; j--) {anss.push_back(s[j]);anst.push_back('-');}}else {for (int j = cj - 1; j >= nj; j--) {anss.push_back('-');anst.push_back(t[j]);}}ci = ni, cj = nj;}reverse(all(anss));reverse(all(anst));cout << dp[n][m] << "\n";cout << anss << "\n";cout << anst << "\n";}signed main() {ios::sync_with_stdio(false);cin.tie(0);cout << fixed << setprecision(10);//init_f();//init();//expr();//while(true)//int t; cin >> t; rep(i, t)solve();return 0;}