結果
問題 | No.709 優勝可能性 |
ユーザー | kou6839 |
提出日時 | 2018-12-23 19:58:15 |
言語 | C++11 (gcc 11.4.0) |
結果 |
AC
|
実行時間 | 525 ms / 3,500 ms |
コード長 | 7,852 bytes |
コンパイル時間 | 1,287 ms |
コンパイル使用メモリ | 128,884 KB |
実行使用メモリ | 14,980 KB |
最終ジャッジ日時 | 2024-09-25 10:36:30 |
合計ジャッジ時間 | 6,232 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge5 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 4 ms
8,192 KB |
testcase_01 | AC | 4 ms
8,320 KB |
testcase_02 | AC | 4 ms
8,192 KB |
testcase_03 | AC | 4 ms
8,064 KB |
testcase_04 | AC | 4 ms
8,192 KB |
testcase_05 | AC | 4 ms
8,192 KB |
testcase_06 | AC | 4 ms
8,064 KB |
testcase_07 | AC | 4 ms
8,064 KB |
testcase_08 | AC | 4 ms
8,064 KB |
testcase_09 | AC | 5 ms
8,192 KB |
testcase_10 | AC | 253 ms
12,544 KB |
testcase_11 | AC | 175 ms
13,892 KB |
testcase_12 | AC | 410 ms
12,544 KB |
testcase_13 | AC | 370 ms
12,416 KB |
testcase_14 | AC | 331 ms
12,544 KB |
testcase_15 | AC | 305 ms
12,416 KB |
testcase_16 | AC | 289 ms
14,276 KB |
testcase_17 | AC | 291 ms
14,980 KB |
testcase_18 | AC | 4 ms
8,192 KB |
testcase_19 | AC | 4 ms
8,192 KB |
testcase_20 | AC | 522 ms
12,160 KB |
testcase_21 | AC | 525 ms
13,600 KB |
testcase_22 | AC | 5 ms
8,064 KB |
testcase_23 | AC | 5 ms
8,064 KB |
ソースコード
#include <vector> #include <list> #include <map> #include <set> #include <queue> #include <deque> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <cctype> #include <string> #include <cstring> #include <ctime> #include <fstream> #include <array> #define _USE_MATH_DEFINES #include <math.h> #include <unordered_set> #include<unordered_map> using namespace std; inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; } template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); } typedef vector<int> vi; typedef vector<vi> vvi; typedef vector<long long int> vll; typedef vector<string> vs; typedef pair<int, int> pii; typedef long long ll; typedef pair<ll, ll> pll; typedef unsigned long long ull; //repetition //------------------------------------------ #define REP(i,a,b) for(int i=(a);i<(b);++i) #define rep(i,n) REP(i,0,n) #define rrep(i,n) for(int i=(n);i>=0;i--) #define VEC_2D(a,b) vector<vector<int> >(a, vector<int>(b, 0)) #define ALL(a) (a).begin(),(a).end() #define RALL(a) (a).rbegin(), (a).rend() #define pb push_back #define mp make_pair #define INF (1001000000) #define SZ(a) int((a).size()) #define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i) #define EXIST(s,e) ((s).find(e)!=(s).end()) #define SORT(c) sort((c).begin(),(c).end()) #define UNIQ(c) (c).erase(unique((c).begin(),(c).end()), (c).end()); #define MOD 1000000007LL #define MS(v,n) memset((v),(n),sizeof(v)) //input #define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i; //output #define P(p) cout<<(p)<<endl; #define FSP(a) cout << fixed << setprecision(a) template<typename T> T gcd(T x, T y) { if (y == 0) return x; else return gcd(y, x%y); } template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; } template<typename T> bool is_prime(T n) { for (int i = 2; i * i <= n; i++) { if (n % i == 0) return false; } return n != 1; } map<ll, ll> prime_factor(ll n) { map<ll, ll> res; for (int i = 2; i * i <= n; i++) { while (n % i == 0) { ++res[i]; n /= i; } } if (n != 1) res[n] = 1; return res; } int extgcd(int a, int b, int& x, int& y) {// int d = a; if (b != 0) { d = extgcd(b, a%b, y, x); y -= (a / b)*x; } else { x = 1; y = 0; } return d; } ll mod_pow(ll x, ll n, ll mod) { if (n == 0) return 1; ll res = mod_pow(x * x % mod, n / 2, mod); if (n & 1) res = res * x % mod; return res; } ll comb(ll a, ll b, ll mod) { ll mul = 1; ll div = 1; rep(i, b) { mul *= (a - (ll)i); mul %= mod; div *= ((ll)i + 1); div %= mod; } mul *= mod_pow(div, mod - 2,mod); return mul%mod; } vector<string> split(const string &str, char delim) { vector<string> res; size_t current = 0, found; while ((found = str.find_first_of(delim, current)) != string::npos) { res.push_back(string(str, current, found - current)); current = found + 1; } res.push_back(string(str, current, str.size() - current)); return res; } bool is_kadomatsu(int a, int b, int c) { if (a == b || a == c || b == c)return false; if (a > b && c > b) return true; if (a < b && c < b)return true; return false; } struct UF { int n; vi d; UF() {} UF(int n) :n(n), d(n, -1) {} int root(int v) { if (d[v] < 0) return v; return d[v] = root(d[v]); } bool same(int a, int b) { return root(a) == root(b); } bool unite(int x, int y) { x = root(x); y = root(y); if (x == y) return false; if (size(x) < size(y)) swap(x, y); d[x] += d[y]; d[y] = x; return true; } int size(int v) { return -d[root(v)]; } }; vector<int> divisor(int n) { if (n == 1) return{}; vi res; for (int i = 1; i*i <= n; i++) { if (n%i == 0) { res.emplace_back(i); if (i != 1 && i != n / i)res.emplace_back(n / i); } } return res; } struct Bellmanford { int n; struct edge { int from, to, cost; }; vector<edge> E; vi d; Bellmanford(int n) :n(n), d(n) { E.resize(n); } void add_edge(int x, int y, int cost) { edge e; e.from = x; e.to = y; e.cost = cost; E.push_back(e); } void shortest_path(int s) { rep(i, n)d[i] = INF; d[s] = 0; while (true) { bool update = false; for (auto e : E) { if (d[e.from] != INF && d[e.to] > d[e.from] + e.cost) { d[e.to] = d[e.from] + e.cost; update = true; } } if (!update) break; } } }; struct Dijkstra { int n; struct edge { int to; ll cost; }; vector<vector<edge>> G; vll d; Dijkstra(int n) :n(n), d(n) { G.resize(n); } void add_edge(int x, int y, ll cost) { edge e; e.to = y; e.cost = cost; G[x].push_back(e); } void shortest_path(int s) { rep(i, n)d[i] = 100000000000000000; d[s] = 0; priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> que; que.push(make_pair(0, s)); while (!que.empty()) { pii p = que.top(); que.pop(); int v = p.second; if (d[v] < p.first) continue; for (auto e : G[v]) { if (d[e.to] > d[v] + e.cost) { d[e.to] = d[v] + e.cost; que.push(make_pair(d[e.to], e.to)); } } } } }; struct Segmenttree { int n; vector<int> dat; Segmenttree(int n_) { n = 1; while (n < n_) n *= 2; dat = vector<int>(2 * n - 1, 0); } void add(int idx, ll val) {//0-indexed idx += n - 1; dat[idx] += val; while (idx > 0) { idx = (idx - 1) / 2; dat[idx] += val; } } int query(int a, int b) { return query_seg(a, b, 0, 0, n); } int query_seg(int a, int b, int k, int l, int r) { if (r <= a || b <= l) return 0; if (a <= l && r <= b)return dat[k]; else { return query_seg(a, b, k * 2 + 1, l, (l + r) / 2) + query_seg(a, b, k * 2 + 2, (l + r) / 2, r); } } }; struct Trie { Trie *next[26]; Trie() { fill(next, next + 26, (Trie *)0); } void insert(char *s) { if (*s == '\0') return; if (this->next[*s - 'a'] == NULL) { this->next[*s - 'a'] = new Trie(); } this->next[*s - 'a']->insert(s + 1); } bool find(char *s) { if (*s == '\0') return true; if (this->next[*s - 'a'] == NULL) { return false; } return this->next[*s - 'a']->find(s + 1); } }; struct edge { int to, cap, rev; }; vector<edge> G[200005]; int level[200005]; int iter[200005]; void add_edge(int from, int to, int cap) { G[from].push_back({ to, cap, (int)G[to].size() }); G[to].push_back({ from, 0, (int)G[from].size() - 1 }); } void fbfs(int s) { memset(level, -1, sizeof(level)); queue<int> que; level[s] = 0; que.push(s); while (!que.empty()) { int v = que.front(); que.pop(); for (edge &e : G[v]) { if (e.cap > 0 && level[e.to] < 0) { level[e.to] = level[v] + 1; que.push(e.to); } } } } int fdfs(int v, int t, int f) { if (v == t) return f; for (edge &e : G[v]) { if (e.cap > 0 && level[v] < level[e.to]) { int d = fdfs(e.to, t, min(f, e.cap)); if (d > 0) { e.cap -= d; G[e.to][e.rev].cap += d; return d; } } } return 0; } int max_flow(int s, int t) { int flow = 0; for (;;) { fbfs(s); if (level[t] < 0) return flow; memset(iter, 0, sizeof(iter)); int f; while ((f = fdfs(s, t, INF)) > 0) { flow += f; } } } //------------------------ int N, M, R[101010][10]; int ma[10]; vi win[10]; int cnt[101010]; int main() { cin >> N >> M; rep(i, N)rep(j, M)cin >> R[i][j]; rep(m, M)ma[m] = R[0][m]; rep(m, M)win[m].pb(0); cnt[0] = M; P(1); int ans = 1; for(int i=1;i<N;i++) { rep(m, M) { if (ma[m] == R[i][m]) { win[m].pb(i); cnt[i]++; if (cnt[i] == 1)ans++; } else if (ma[m] < R[i][m]){ for (int j : win[m]) { cnt[j]--; if (cnt[j] == 0)ans--; } win[m].clear(); ma[m] = R[i][m]; win[m].pb(i); cnt[i]++; if (cnt[i] == 1)ans++; } } P(ans); } return 0; }