結果

問題 No.861 ケーキカット
コンテスト
ユーザー zengyongxu-jerry
提出日時 2026-06-21 22:23:21
言語 C++23
(gcc 15.2.0 + boost 1.89.0)
コンパイル:
g++-15 -O2 -lm -std=c++23 -Wuninitialized -DONLINE_JUDGE -o a.out _filename_
実行:
./a.out
結果
WA  
実行時間 -
コード長 5,286 bytes
記録
記録タグの例:
初AC ショートコード 純ショートコード 純主流ショートコード 最速実行時間
コンパイル時間 3,655 ms
コンパイル使用メモリ 347,332 KB
実行使用メモリ 6,400 KB
最終ジャッジ日時 2026-06-21 22:23:34
合計ジャッジ時間 5,013 ms
ジャッジサーバーID
（参考情報） 		judge1_0 / judge2_0
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 * 4 WA * 17
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 #
raw source code 

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define int long long
#define pii pair<int, int>
#define piii pair<pii, int>
#define pll pair<ll, ll>
#define plll pair<pll, ll>
#define fi first
#define se second
const int N = 2e5 + 5, M = 1e6 + 5;
const int inf = 1e9, mod = 998244353;
const ll INF = 1e18;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
 
namespace ARIS0_0{
	int a[10][10], id[10][10];
	
	void init(){
	}
	void solve(){
		int n = 5, all = 0;
		for (int i = 1; i <= 5; i ++ )
			for (int j = 1; j <= 5; j ++ )
				cin >> a[i][j], all += a[i][j];
		for (int i = 1, t = 1; i <= 5; i ++ )
			for (int j = 1; j <= 5; j ++, t ++ )
				id[i][j] = t;
		
		int ans = INF;
		for (int I = 0; I < 16; I ++ )
			for (int J = I; J < 16; J ++ ){
				int sx, sy, ex, ey;
				{
					int a = I / 4, b = I % 4;
					if (a == 0) sx = 1, sy = b + 1;
					else if (a == 1) sx = b + 1, sy = n;
					else if (a == 2) sx = n, sy = n - b;
					else if (a == 3) sx = n - b, sy = 1;
				}
				{
					int a = J / 4, b = J % 4;
					if (a == 0) ex = 1, ey = b;
					else if (a == 1) ex = b, ey = n;
					else if (a == 2) ex = n, ey = n - b + 1;
					else if (a == 3) ex = n - b + 1, ey = 1;
				}
				int sa = 0;
				
				vector <vector <int>> vis(6, vector <int> (6));
				if (I / 4 == J / 4){
					int aa = I / 4;
					if (aa == 0){
						for (int j = sy; j <= ey; j ++ ) sa += a[1][j], vis[1][j] = 1;
					}
					else if (aa == 1){
						for (int j = sx; j <= ex; j ++ ) sa += a[j][n], vis[j][n] = 1;
					}
					else if (aa == 2){
						for (int j = sy; j >= ey; j -- ) sa += a[n][j], vis[n][j] = 1;
					}
					else{
						for (int j = sx; j >= ex; j -- ) sa += a[j][1], vis[j][1] = 1;
					}
				}
				else{
					{
						int aa = I / 4, b = I % 4;
						if (aa == 0){
							for (int j = sy; j <= n; j ++ ) if (!vis[1][j]) sa += a[1][j], vis[1][j] = 1;
						}
						else if (aa == 1){
							for (int j = sx; j <= n; j ++ ) if (!vis[j][n]) sa += a[j][n], vis[j][n] = 1;
						}
						else if (aa == 2){
							for (int j = sy; j >= 1; j -- ) if (!vis[n][j]) sa += a[n][j], vis[n][j] = 1;
						}
						else if (aa == 3){
							for (int j = sx; j >= 1; j -- ) if (!vis[n][j]) sa += a[j][1], vis[j][1] = 1;
						}
					}
					{
						int aa = J / 4, b = J % 4;
						if (aa == 0){
							for (int j = 1; j <= sy; j ++ ) if (!vis[1][j]) sa += a[1][j], vis[1][j] = 1;
						}
						else if (aa == 1){
							for (int j = 1; j <= sx; j ++ ) if (!vis[j][n]) sa += a[j][n], vis[j][n] = 1;
						}
						else if (aa == 2){
							for (int j = n; j >= sy; j -- ) if (!vis[n][j]) sa += a[n][j], vis[n][j] = 1;
						}
						else if (aa == 3){
							for (int j = n; j >= sx; j -- ) if (!vis[n][j]) sa += a[j][1], vis[j][1] = 1;
						}
					}
					for (int aa = I / 4 + 1; aa < J / 4; aa ++ ){
						if (aa == 0){
							for (int j = 1; j <= n; j ++ ) if (!vis[1][j]) sa += a[1][j], vis[1][j] = 1;
						}
						else if (aa == 1){
							for (int j = 1; j <= n; j ++ ) if (!vis[j][n]) sa += a[j][n], vis[j][n] = 1;
						}
						else if (aa == 2){
							for (int j = n; j >= 1; j -- ) if (!vis[n][j]) sa += a[n][j], vis[n][j] = 1;
						}
						else if (aa == 3){
							for (int j = n; j >= 1; j -- ) if (!vis[j][1]) sa += a[j][1], vis[j][1] = 1;
						}
					}
				}
				
				for (int m = 0; m < (1 << 9); m ++ ){
					for (int j = 0; j < 9; j ++ )
						if (((m >> j) & 1)){
							int x = j / 3, y = j % 3;
							x ++, y ++;
							vis[x][y] = 1;
						}
					
					int cnt1 = 0;
					queue <pii> q;
					q.push({sx, sy});
					vector <vector <int>> qwq(6, vector <int> (6));
					while (q.size()){
						int x = q.front().fi, y = q.front().se; q.pop();
						if (qwq[x][y]) continue;
						qwq[x][y] = 1, cnt1 ++ ;
						for (int u = 0; u < 4; u ++ ){
							int nx = x + dx[u], ny = y + dy[u];
							if (nx < 1 || nx >n || ny < 1 || ny > n) continue;
							if (vis[nx][ny]) q.push({nx, ny});
						}
					}
					
					int cnt2 = 0;
					while (q.size()) q.pop();
					for (int i = 1; i <= n; i ++ )
						for (int j = 1; j <= n; j ++ ){
							if (!vis[i][j] && !q.size())
								q.push({i, j});
							qwq[i][j] = false;
						}
					while (q.size()){
						int x = q.front().fi, y = q.front().se; q.pop();
						if (qwq[x][y]) continue;
						qwq[x][y] = 1, cnt2 ++ ;
						for (int u = 0; u < 4; u ++ ){
							int nx = x + dx[u], ny = y + dy[u];
							if (nx < 1 || nx >n || ny < 1 || ny > n) continue;
							if (!vis[nx][ny]) q.push({nx, ny});
						}
					}
					if (cnt1 + cnt2 == 25){
						int x = abs(sa - (all - sa));
						ans = min(ans, x);
					}
				}
			}
		cout << ans << "\n";
	}
	void single(){ init(), solve(); }
	void multi(){ init(); int T; cin >> T; while (T -- ) solve(); }
	void idmulti(){ init(); int id, T; cin >> id >> T; while (T -- ) solve(); }
};
 
signed main(){
//	freopen("cake.in", "r", stdin);
//	freopen("cake.out", "w", stdout);
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	ARIS0_0::single();
}

/*

注意到最外圈有两个或零个切口
枚举这两个，假设中间某一段是 A，另一段是 B
然后向中间扩张
中间只有 3*3！！！
于是可以 2^9 枚举！！！
然后判断联通即可！
联通！
fish_love_dsu！！！ 
靠怎么变量名重名了，i -> I, j -> J 得了 

*/
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