Yahoo Programming Contest 2019 D - Ears

D - Ears

链接

分析：

　　转化一下题意，将一段序列分成5段，空 偶 奇 偶 空，然后dp一下，f[i][j]表示到第i个数，当前在第j段内，的最小花费。

　　空段中的花费是数字大小，偶数段中奇数花费1，奇数段中偶数花费1。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 200005;
int a[N];
LL f[N][5];

int main() {
    int n = read();
    for (int i = 1; i <= n; ++i) a[i] = read();
    for (int i = 1; i <= n; ++i) {
        f[i][0] = f[i - 1][0];
        for (int j = 1; j <= 4; ++j) f[i][j] = min(f[i - 1][j], f[i][j - 1]);
        f[i][0] += a[i];
        f[i][1] += (a[i] == 0) ? 2 : (a[i] & 1);
        f[i][2] += !(a[i] & 1);
        f[i][3] += (a[i] == 0) ? 2 : (a[i] & 1);
        f[i][4] += a[i];
    }
    LL ans = f[n][0];
    for (int i = 1; i <= 4; ++i) ans = min(ans, f[n][i]);
    cout << ans;
    return 0;
}