Takahashi and Animals

https://atcoder.jp/contests/abc251/tasks/abc251_e

Solution

參考

https://blog.csdn.net/qq_52678569/article/details/124790849

https://blog.csdn.net/qq_34364611/article/details/124784187

Code

#include <bits/stdc++.h>
#include <vector>
#include <algorithm>
#include <deque>
#include 
 
<queue>
#include <string>
#include <set>
using namespace std;


/*
https://atcoder.jp/contests/abc251/tasks/abc251_e
*/

const int N = 300000 + 16;

int n;
int a[N] = {0};

long long dp[N][2] = {{0}};

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }

    
 
/*
    dp[i][j] -- the cost to feed all top i cats
    but for i cat, it has two situations:
    dp[i][0] -- don't feed i cat with a[i] food cost
    dp[i][1] -- feed i cat with a[i] food cost

    generally speaking:
    dp[i][0] = dp[i-1][1]  // due to lack of feeding a[i] to i cat, then i-1 cat must be fed
    dp[i][1] = min(dp[i-1][0], dp[i-1][1]) + a[i] // due to i cat fed with a[i], so i-1 cat can be fed or not

    but for i == 1, it depends on if n cat was fed with a[n] or not
    case 1
        if n cat was fed with a[n], then 1 cat can be fed or not
        dp[1][0] = 0
        dp[1][1] = a[1]
    case 2
        if n cat was not fed with a[n], then 1 cat should be fed only
        dp[1][0] = INF
        dp[1][1] = a[1]
    
 
*/

    long long ans = 1e18;

    // for case 1
    dp[1][0] = 0;
    dp[1][1] = a[1];

    for (int i = 2; i <= n; i++) {
        dp[i][0] = dp[i - 1][1];
        dp[i][1] = min(dp[i - 1][0], dp[i - 1][1]) + a[i];
    }

    ans = min({ans, dp[n][1]});

    // for case 2
    dp[1][0] = 1e18;
    dp[1][1] = a[1];

    for (int i = 2; i <= n; i++) {
        dp[i][0] = dp[i - 1][1];
        dp[i][1] = min(dp[i - 1][0], dp[i - 1][1]) + a[i];
    }

    ans = min({ans, dp[n][0]});

    cout << ans << endl;
}