Ural 2072

題意：澆花，必須以權值從小到大澆水，左右走一步需要一秒，澆完所有的花最少需要多少時間。
思路：當走到某一步時，發現左右都有權值相同的，這就要考慮往左走還是往右走，所以記憶化處理一下，每次取最小值。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;
const int maxn = 100007;
int n;

struct Node {
    LL val, id;
    bool operator < (const Node& rhs) const {
        return val < rhs.val || (val == rhs.val && id < rhs.id);
    }
}p[maxn];

int main() {
    scanf("%d", &n);
    for(int i = 0; i < n; ++i) scanf("%lld", &p[i].val), p[i].id = i;
    sort(p, p+n);
    LL prel = 0, prer = 0;
    LL ansl = 0, ansr = 0;
    for(int i = 0; i < n; ++i) {
        int j, cnt = i;
        for(j = i+1; j < n; ++j) if(p[i].val != p[j].val) break;
        cnt = j-1;

        LL l = p[i].id, r = p[cnt].id;

        LL aansl = abs(l-r) + min(abs(r-prel)+ansl, abs(r-prer)+ansr);
        LL aansr = abs(l-r) + min(abs(l-prel)+ansl, abs(l-prer)+ansr);


        prel = l; prer = r;
        ansl = aansl; ansr = aansr;
        i = cnt;
    }
    printf("%lld\n", min(ansl, ansr)+n);
}