Submission #73114403

Source Code Expand

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#include <stdio.h>
int N;
int iro[11234];
int rle_num;
int rle_color[11234], rle_cnt[11234];
int kesu(int up, int down) {
  int kieta = 0;
  while (rle_color[up] == rle_color[down] && rle_cnt[up] + rle_cnt[down] >= 4) {
    kieta += rle_cnt[up] + rle_cnt[down];
    up--;
    down++;
  }
  return kieta;
}
int main(void) {
  int i;
  int kesu_max = 0, candidate;
 
 
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#include <stdio.h>

int N;
int iro[11234];

int rle_num;
int rle_color[11234], rle_cnt[11234];

int kesu(int up, int down) {
	int kieta = 0;
	while (rle_color[up] == rle_color[down] && rle_cnt[up] + rle_cnt[down] >= 4) {
		kieta += rle_cnt[up] + rle_cnt[down];
		up--;
		down++;
	}
	return kieta;
}

int main(void) {
	int i;
	int kesu_max = 0, candidate;
	if (scanf("%d", &N) != 1) return 1;
	for (i = 0; i < N; i++) {
		if (scanf("%d", &iro[i]) != 1) return 1;
	}

	rle_num = 2;
	rle_color[0] = -1;
	rle_cnt[0] = 0;
	rle_color[1] = iro[0];
	rle_cnt[1] = 1;
	for (i = 1; i < N; i++) {
		if (rle_color[rle_num - 1] == iro[i]) {
			rle_cnt[rle_num - 1]++;
		} else {
			rle_color[rle_num] = iro[i];
			rle_cnt[rle_num] = 1;
			rle_num++;
		}
	}
	rle_color[rle_num] = -1;
	rle_cnt[rle_num] = 0;
	rle_num++;

	for (i = 1; i < rle_num - 1; i++) {
		if (rle_cnt[i] > 1) {
			/* このブロックは2個以上 */
			rle_cnt[i]--;
			if (rle_cnt[i - 1] == 3) {
				/* 上のブロックにくっつける */
				candidate = 4 + kesu(i - 2, i);
				if (candidate > kesu_max) kesu_max = candidate;
			}
			if (rle_cnt[i + 1] == 3) {
				/* 下のブロックにくっつける */
				candidate = 4 + kesu(i, i + 2);
				if (candidate > kesu_max) kesu_max = candidate;
			}
			rle_cnt[i]++;
		} else {
			/* このブロックは1個 */
			if (rle_color[i - 1] == rle_color[i + 1]) {
				/* 上下のブロックは色が同じ */
				int naka = rle_cnt[i - 1] + rle_cnt[i] + rle_cnt[i + 1];
				if (naka >= 4) {
					candidate = naka + kesu(i - 2, i + 2);
					if (candidate > kesu_max) kesu_max = candidate;
				}
			} else {
				/* 上下のブロックは色が違う */
				if (rle_cnt[i - 1] == 3) {
					/* 上のブロックにくっつける */
					candidate = 4 + kesu(i - 2, i + 1);
					if (candidate > kesu_max) kesu_max = candidate;
				}
				if (rle_cnt[i + 1] == 3) {
					/* 下のブロックにくっつける */
					candidate = 4 + kesu(i - 1, i + 2);
					if (candidate > kesu_max) kesu_max = candidate;
				}
			}
		}
	}
	printf("%d\n", N - kesu_max);
	return 0;
}

/*

1回消すと4個以上消費するので、連鎖の回数は高々2500回
N=2500のO(N^2) ならいけるはず
今回は 10000×2500 だけど、制限時間10秒あるからいけるやろ

*/

Submission Info

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