Yahoo Programming Contest 2019 D - Ears

D - Ears

思路:

Yahoo Programming Contest 2019 D - Ears

s:起点           t:终点           l:左端点           r:右端点

以上称为关键点

dp[i][j]表示到位置 i 为止,已经经过前 j 个关键点的最小花费

转移方程看代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head const int N = 2e5 + ;
LL dp[N][];
int a[N];
int main() {
int n;
scanf("%d", &n);
for (int i = ; i <= n; ++i) scanf("%d", &a[i]);
for (int i = ; i <= n; i++) {
dp[i][] = dp[i-][] + a[i];
for (int j = ; j < ; j++) {
dp[i][j] = dp[i-][j];
for (int k = ; k < j; k++) dp[i][j] = min(dp[i][j], dp[i-][k]);
}
dp[i][] += a[i] == ? : (a[i]% != );
dp[i][] += a[i]% != ;
dp[i][] += a[i] == ? : (a[i]% != );
dp[i][] += a[i];
}
LL ans = dp[n][];
for (int i = ; i < ; i++) ans = min(ans, dp[n][i]);
printf("%lld\n", ans);
return ;
}
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