Hills And Valleys

标签：影响   int   细节   ble   lowbit   cond   c++   return   ORC   

http://codeforces.com/contest/1467/problem/B

题意

\(一个数列里有波峰波谷\)

• \(波谷:a[i]<min(a[i+1],a[i-1]])\)
• \(波峰:a[i]>max(a[i+1],a[i-1])\)
• \(sum = 波峰数+波谷数\)

\(若可以任意修改一个a[i]的值,问如何修改使得序列的sum值最小.\)

思路

\(如果去想修改的细节会很麻烦,所以要找到能够遍历解决的通法\\对于修改任意一个数a[i],其值只会影响到a[i-1]和a[i+1]是否为波峰波谷,\\所以每枚举一个i只用考虑其三元组的变化.\)

代码

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define fi first
#define se second
#define inf 0x3f3f3f3f
const int N = 3e5 + 7;
int a[N];
int n;

int cal(int i) {
    if (i <= 1 || i >= n) return 0;
    int res = 0;
    if (a[i] < min(a[i + 1], a[i - 1])) ++res;
    if (a[i] > max(a[i + 1], a[i - 1])) ++res;
    return res;
}

int main() {
    IO;
    int _;
    cin >> _;
    while (_--) {
        cin >> n;
        for (int i = 1; i <= n; ++i) cin >> a[i];
        int sum = 0, ans = inf;
        for (int i = 1; i <= n; ++i) sum += cal(i);
        for (int i = 1; i <= n; ++i) {
            int tmp = sum - cal(i) - cal(i - 1) - cal(i + 1);
            int t = a[i];
            a[i] = a[i - 1];
            ans = min(ans, tmp + cal(i) + cal(i - 1) + cal(i + 1));
            a[i] = a[i + 1];
            ans = min(ans, tmp + cal(i) + cal(i - 1) + cal(i + 1));
            a[i] = t;
        } 
        cout << ans << ‘\n‘;
    }
    return 0;
}

Hills And Valleys

标签：影响   int   细节   ble   lowbit   cond   c++   return   ORC   

原文地址：https://www.cnblogs.com/phr2000/p/14391456.html