题解:
dp[i][j][0]:前i天已经j天不穿拖鞋且第i天穿了拖鞋的最大值
dp[i][j][1]:前i天已经j天不穿拖鞋且第i天没穿拖鞋的最大值
状态转移:
dp[i][j][0]=max(dp[i-1][j][0],dp[i-1][j][1]);
dp[i][j][1]=max(dp[i][j][1],dp[i-1][j-1][0]+d[i]);(d[i]表示第i天有多少个妹子)
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<vector>
#include<string>
#include<math.h>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long ll;
#define inf 1000000000
#define mod 1000000007
#define maxn 1705
#define PI acos(-1.0)
#define lowbit(x) (x&-x)
#define eps 1e-9
int dp[maxn][maxn][2], a[maxn*100], b[maxn*100], c[maxn*100], d[maxn*100];
int main(void)
{
int n, k, i, j, cnt, p;
while (scanf("%d%d", &n, &k) != EOF)
{
cnt = 0;p = 1;
memset(a, 0, sizeof(a));
memset(c, 0, sizeof(c));
memset(d, 0, sizeof(d));
memset(dp, 0, sizeof(dp));
for (i = 1;i <= n;i++)
scanf("%d", &a[i]), b[i] = a[i];
k = min(k, n);
sort(b + 1, b + n + 1);
int num = unique(b + 1, b + n + 1) - b - 1;
c[1] = b[1];cnt = 1;
for (i = 2;i <= num;i++)
{
if (b[i] - b[i - 1] > 1)
c[++cnt] = b[i] - 1;
c[++cnt] = b[i];
}
for (i = 1;i <= n;i++)
a[i] = lower_bound(c + 1, c + cnt + 1, a[i]) - c;
sort(a + 1, a + n + 1);
for (i = 1;i <= max(n, cnt);i++)
d[a[i]]++;
for (i = 1;i <= max(cnt, n);i++)
{
for (j = 1;j <= k;j++)
{
if (j > i)
break;
dp[i][j][0] = max(dp[i - 1][j][0], dp[i - 1][j][1]);
dp[i][j][1] = max(dp[i][j][1], dp[i - 1][j - 1][0] + d[i]);
}
}
int ans=0, tmp;
for (i = 0;i <= k;i++)
{
tmp = max(dp[max(n, cnt)][i][1], dp[max(n, cnt)][i][0]);
ans = max(ans, tmp);
}
printf("%d/n", ans);
}
return 0;
}
原创文章,作者:kepupublish,如若转载,请注明出处:https://blog.ytso.com/140469.html