题意:
给定N≤2000的两个序列,通过交换第一个序列变成第二个序列
如果交换ai和aj,交换的代码为|i−j|,给出交换代价最小的一个构造方案
分析:
显然对于每次交换,交换的双方必然向着它们该去的方向交换是最优的,显然这样交换是正确方案之一
发现这个之后如何构造一个解
考虑从右往左构造,如果1要到2,直接换过去显然是可以的
但是中途我们发现3,4要去1那边,显然我们将1和3,4交换,达成了共赢
这样就构造出了一个交换方案了
代码:
//// Created by TaoSama on -01-28// Copyright (c) TaoSama. All rights reserved.//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << " "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, a[N], b[N];int to[N];int main() {#ifdef LOCALfreopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endifios_base::sync_with_stdio(0);while(scanf("%d", &n) == 1) {for(int i = 1; i <= n; ++i) scanf("%d", a + i);for(int i = 1; i <= n; ++i) scanf("%d", b + i);for(int i = 1; i <= n; ++i) to[b[i]] = i;int ans = 0;vector<pair<int, int> > path;for(int i = n; i; --i) {int pos = i;for(int j = i + 1; j <= to[a[pos]]; ++j) {if(to[a[j]] <= pos) {ans += abs(pos - j);path.push_back({pos, j});swap(a[pos], a[j]);pos = j;}}}printf("%d\n%d\n", ans, path.size());for(auto p : path) printf("%d %d\n", p.first, p.second);}return 0;}
