Codeforces Round #743 (Div. 2)

B. Swaps

思路：

把a，b数组开成pair，first存值，second存下标，然后把两个数组排序，由于在同一位置下，a数组永远小于b数组，所以我们每到一个位置，就取一个min的dist（a[i]到开头的距离），保证a[i]的位置尽量靠前，然后再取一个min的res，由于下标从1开始，所以res取最小的b.second + dist - 2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;

typedef pair<int, int>PII;
typedef long long LL;

const int N = 100010;

PII a[N], b[N];

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n;
        cin >> n;
        for (int i = 1; i <= n; i++) cin >> a[i].first, a[i].second = i;
        for (int i = 1; i <= n; i++) cin >> b[i].first, b[i].second = i;

        sort(a + 1, a + 1 + n);
        sort(b + 1, b + 1 + n);

        int dist = 0x3f3f3f3f, res = 0x3f3f3f3f;
        for (int i = 1; i <= n; i++)
        {
            dist = min(dist, a[i].second);//随着i的增加，a[i]要保证位置要尽量靠前
            res = min(res, b[i].second + dist - 2);
        }

        cout << res << endl;
    }
    return 0;

}

//3(1) 1(2)
//4(1) 2(2)
//
//1(2) 3(1)
//2(2) 4(1)
//
//7(1) 5(2) 9(3) 1(4) 3(5)
//2(1) 4(2) 6(3) 10(4) 8(5)
//
//1(4) 3(5) 5(2) 7(1) 9(3)
//2(1) 4(2) 6(3) 8(5) 10(4)

标签：743,dist,int,res,Codeforces,second,Div,include,first
来源： https://www.cnblogs.com/yctql/p/15313844.html