提出 #77963303

ソースコード 拡げる

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#include <stdio.h>
#include <inttypes.h>
int N, K;
int A[212345], B[212345];
uint64_t memo[212345][16][2];
uint64_t calc(int pos, int left, int flipped) {
  uint64_t ans = 0;
  if (pos >= N) return 0;
  if (memo[pos][left][flipped]) return ~memo[pos][left][flipped];
  /* 表面を使う */
  ans = calc(pos + 1, left, 0) + A[pos];
  /* 裏面を使う */
  if (left > 0 || flipped) {
    uint64_t candidate = calc(pos + 1, left - !flipped, 1) + B[pos];
    if (candidate > ans) ans = candidate;
  }
 
 
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#include <stdio.h>
#include <inttypes.h>

int N, K;
int A[212345], B[212345];

uint64_t memo[212345][16][2];

uint64_t calc(int pos, int left, int flipped) {
	uint64_t ans = 0;
	if (pos >= N) return 0;
	if (memo[pos][left][flipped]) return ~memo[pos][left][flipped];

	/* 表面を使う */
	ans = calc(pos + 1, left, 0) + A[pos];
	/* 裏面を使う */
	if (left > 0 || flipped) {
		uint64_t candidate = calc(pos + 1, left - !flipped, 1) + B[pos];
		if (candidate > ans) ans = candidate;
	}

	return ~(memo[pos][left][flipped] = ~ans);
}

int main(void) {
	int i;
	if (scanf("%d%d", &N, &K) != 2) return 1;
	for (i = 0; i < N; i++) {
		if (scanf("%d%d", &A[i], &B[i]) != 2) return 1;
	}
	printf("%" PRIu64 "\n", calc(0, K, 0));
	return 0;
}

/*

    -----
  ---------
* * * * * * *

を

  -       -
* * * * * * *

    -------
  -----
* * * * * * *

を

  -     ---
* * * * * * *

みたいにできるので、多分同じカードを複数回裏返す必要はない

*/

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