제출 #405669

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
405669		rainboy	Paint By Numbers (IOI16_paint)	C11	100 / 100	323 ms	42444 KiB

paint

#include "paint_c.h"

#define N	200000
#define K	100

void solve_puzzle(int n, char *aa, int k, int *ll, char *cc) {
	static char dp[N + 1][K + 1], dq[N + 1][K + 1], can_w[N + 1];
	static int pp[N], can_b[N + 1];
	int h, i, l;

	for (i = 0; i < n; i++)
		pp[i] = aa[i] == '_' ? i : (i == 0 ? -1 : pp[i - 1]);
	dp[0][0] = 1;
	for (i = 1; i <= n; i++)
		for (h = 0; h <= k; h++) {
			l = h == 0 ? -1 : ll[h - 1];
			dp[i][h] = aa[i - 1] != 'X' && dp[i - 1][h] || h > 0 && pp[i - 1] < i - l && (i == l ? h == 1 : aa[i - 1 - l] != 'X' && dp[i - 1 - l][h - 1]);
		}
	dq[n][k] = 1;
	for (i = n - 1; i >= 0; i--)
		for (h = k; h >= 0; h--) {
			l = h == k ? -1 : ll[h];
			dq[i][h] = aa[i] != 'X' && dq[i + 1][h] || h < k && i + l <= n && pp[i + l - 1] < i && (i + l == n ? h == k - 1 : aa[i + l] != 'X' && dq[i + l + 1][h + 1]);
		}
	for (i = 0; i < n; i++)
		for (h = 0; h <= k; h++)
			if (dp[i][h] && dq[i + 1][h]) {
				can_w[i] = 1;
				break;
			}
	for (h = 0; h < k; h++) {
		l = ll[h];
		for (i = 0; i + l <= n; i++)
			if (pp[i + l - 1] < i && (i == 0 ? h == 0 : aa[i - 1] != 'X' && dp[i - 1][h]) && (i + l == n ? h == k - 1 : aa[i + l] != 'X' && dq[i + l + 1][h + 1]))
				can_b[i]++, can_b[i + l]--;
	}
	for (i = 1; i <= n; i++)
		can_b[i] += can_b[i - 1];
	for (i = 0; i < n; i++)
		cc[i] = aa[i] == '.' ? (can_w[i] && can_b[i] ? '?' : (can_w[i] ? '_' : 'X')) : aa[i];
}

컴파일 시 표준 에러 (stderr) 메시지

paint.c: In function 'solve_puzzle':
paint.c:17:32: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
   17 |    dp[i][h] = aa[i - 1] != 'X' && dp[i - 1][h] || h > 0 && pp[i - 1] < i - l && (i == l ? h == 1 : aa[i - 1 - l] != 'X' && dp[i - 1 - l][h - 1]);
      |               ~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~
paint.c:23:28: warning: suggest parentheses around '&&' within '||' [-Wparentheses]
   23 |    dq[i][h] = aa[i] != 'X' && dq[i + 1][h] || h < k && i + l <= n && pp[i + l - 1] < i && (i + l == n ? h == k - 1 : aa[i + l] != 'X' && dq[i + l + 1][h + 1]);
      |               ~~~~~~~~~~~~~^~~~~~~~~~~~~~~

• Subtask 1: 7 / 7
• Subtask 2: 3 / 3
• Subtask 3: 22 / 22
• Subtask 4: 27 / 27
• Subtask 5: 21 / 21
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10