哈尔滨市建设安全网站,网站开发账务处理,湖北百度推广公司,wordpress扒皮传送门 文章目录题意#xff1a;思路#xff1a;题意#xff1a;
给你两个序列a,ba,ba,b#xff0c;每次可以执行一个操作#xff1a;将a[i]a[i]a[i]与a[i1]a[i1]a[i1]交换#xff0c;且让交换后的a[i]1,a[i1]−1a[i]1,a[i1]-1a[i]1,a[i1]−1#xff0c;问将aaa变成bbb…传送门
文章目录题意思路题意
给你两个序列a,ba,ba,b每次可以执行一个操作将a[i]a[i]a[i]与a[i1]a[i1]a[i1]交换且让交换后的a[i]1,a[i1]−1a[i]1,a[i1]-1a[i]1,a[i1]−1问将aaa变成bbb的最小操作数。不能变成的话输出−1-1−1。
思路
考虑将aiaii,bibiia_ia_ii,b_ib_iiaiaii,bibii如果有解的话那么这两个序列必须相同。证明不是很明白待补。 之后就是裸题了用线段树维护一下位置求出来就好了。
// Problem: E. String Reversal
// Contest: Codeforces - Educational Codeforces Round 96 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1430/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize(Ofast,no-stack-protector,unroll-loops,fast-math)
//#pragma GCC target(sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tunenative)
//#pragma GCC optimize(2)
#includecstdio
#includeiostream
#includestring
#includecstring
#includemap
#includecmath
#includecctype
#includevector
#includeset
#includequeue
#includealgorithm
#includesstream
#includectime
#includecstdlib
#define X first
#define Y second
#define L (u1)
#define R (u1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].ltr[u].r1)
#define Len(u) (tr[u].r-tr[u].l1)
#define random(a,b) ((a)rand()%((b)-(a)1))
#define db puts(---)
using namespace std;//void rd_cre() { freopen(d://dp//data.txt,w,stdout); srand(time(NULL)); }
//void rd_ac() { freopen(d://dp//data.txt,r,stdin); freopen(d://dp//AC.txt,w,stdout); }
//void rd_wa() { freopen(d://dp//data.txt,r,stdin); freopen(d://dp//WA.txt,w,stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pairint,int PII;const int N1000010,mod1e97,INF0x3f3f3f3f;
const double eps1e-6;int n;
int a[N],b[N];
mapint,vectorintv;
struct Node {int l,r;LL cnt,lazy;
}tr[N2];void pushdown(int u) {if(!tr[u].lazy) return;LL lazytr[u].lazy; tr[u].lazy0;tr[L].lazylazy; tr[R].lazylazy;tr[L].cntlazy; tr[R].cntlazy;
}void build(int u,int l,int r) {tr[u]{l,r,1,0};if(lr) {tr[u].cntl;return;}build(L,l,Mid); build(R,Mid1,r);
} void change(int u,int l,int r,int c) {if(tr[u].lltr[u].rr) {tr[u].cntc;tr[u].lazyc;return;}pushdown(u);if(lMid) change(L,l,r,c);if(rMid) change(R,l,r,c);
}int query(int u,int l,int r) {if(tr[u].lltr[u].rr) return tr[u].cnt;int ans0;pushdown(u);if(lMid) ansquery(L,l,r);if(rMid) ansquery(R,l,r);return ans;
}LL solve() {LL ans0;build(1,1,n);for(int i1;in;i) {int posv[b[i]].back(); v[b[i]].pop_back();LL nowquery(1,pos,pos);ansnow-i;change(1,1,pos,1);}return ans;
}bool check() {mapint,intmp1,mp2;for(int i1;in;i) mp1[a[i]];for(int i1;in;i) mp2[b[i]];for(int i1;in;i) if(mp1[a[i]]!mp2[a[i]]) return false;return true;
}int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);int _1;while(_--) {cinn;for(int i1;in;i) scanf(%d,a[i]),a[i]i;for(int i1;in;i) scanf(%d,b[i]),b[i]i;for(int in;i1;i--) v[a[i]].pb(i);if(!check()) {puts(-1);return 0;}printf(%lld\n,solve());}return 0;
}
/**/
