結果

問題 No.2982 Logic Battle
ユーザー The Forsaking
提出日時 2024-12-25 15:53:39
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 960 ms / 2,000 ms
コード長 3,458 bytes
コンパイル時間 1,300 ms
コンパイル使用メモリ 117,080 KB
最終ジャッジ日時 2025-02-26 16:40:44
ジャッジサーバーID
（参考情報） 		judge1 / judge1
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
other AC * 38
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:89:42: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   89 |     for (int i = 1; i < n + 1; i++) scanf("%d%d%d", a[i], a[i] + 1, a[i] + 2);
      |                                     ~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include <iostream>
#include <sstream>
#include <iomanip>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <vector>
#include <queue> 
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <ctime>
#include <assert.h>
#include <deque>
#include <list>
#include <stack>
using namespace std;
#define is_mul_overflow(a, b) \
    ((b != 0) && (a > LLONG_MAX / b || a < LLONG_MIN / b))
 
typedef pair<long long, int> pli;
typedef pair<int, long long> pil;
typedef pair<long long , long long> pll;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef pair<int, pii> piii;
typedef pair<int, long long > pil;
typedef pair<long long, pii> plii;
typedef pair<double, int> pdi;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ull, ull> puu;
typedef long double ld;
const int N = 2000086, MOD = 998244353, INF = 0x3f3f3f3f, MID = 333;
const long double EPS = 1e-8;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
// int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
// int dx[8] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};
int n, m, cnt;
int w[N];
vector<ll> num;
ll res;
ll lowbit(ll x) { return x & -x; }
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
inline double rand(double l, double r) { return (double)rand() / RAND_MAX * (r - l) + l; }
inline ll qmi(ll a, ll b, ll c) { ll res = 1; while (b) { if (b & 1) res = res * a % c; a = a * a % c; b >>= 1; } return res; }
inline ll qmi(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; }
inline double qmi(double a, ll b) { double res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; } 
// inline ll C(ll a, ll b) { if (a < b) return 0; if (b > a - b) b = a - b; ll res = 1; for (ll i = 1, j = a; i <= b; i++, j--) { res = res * (j % MOD
    ) % MOD; res = res * qmi(i, MOD - 2, MOD) % MOD; } return res; }
inline ll C(ll a, ll b, int* c) { if (a < b) return 0; ll res = 1; for (ll j = a, i = 1; i < b + 1; i++, j--) res *= j; for (ll j = a, i = 1; i < b + 
    1; i++, j--) res /= i; return res; }
inline int find_(int x) { return lower_bound(num.begin(), num.end(), x) - num.begin(); }
int a[5086][3];
ll f[2][5086][3];
inline void solve() {
    memset(f[0], -0x3f, sizeof f[0]);
    f[0][0][0] = f[0][0][1] = f[0][0][2] = 0;
    for (int i = 1; i < n + 1; i++) {
        memset(f[i & 1], -0x3f, sizeof f[i & 1]);
        for (int j = 0; j <= n; j++)
            for (int k = 0; k < 3; k++)
                for (int l = 0; l < 3; l++)
                    if (k != l) {
                        int lv = max(0, j - 1) + 1, rv = min(n - i, lv - 1 + a[i][l]);
                        ll t = 0;
                        if (lv - 1 + a[i][l] > n - i) t = (ll)min(a[i][l], lv - 1 + a[i][l] - (n - i)) * (n - i + 1);
                        if (rv >= lv) t += (ll)(lv + rv) * (rv - lv + 1) / 2;
                        f[i & 1][min(n, max(0, j - 1) + a[i][l])][l] = max(f[i & 1][min(n, max(0, j - 1) + a[i][l])][l], f[i & 1 ^ 1][j][k] + t);
                    }
    }
    for (int i = 0; i < n + 1; i++) res = max({res, f[n & 1][i][0], f[n & 1][i][1], f[n & 1][i][2]});
    printf("%lld\n", res);
}
int main() {
    cin >> n;
    for (int i = 1; i < n + 1; i++) scanf("%d%d%d", a[i], a[i] + 1, a[i] + 2);
    solve();
    return 0;
}
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