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題意：

思路：

與上一篇題解大同小異，無非就是不需要枚舉排列了。

// 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,tune=native") //#pragma GCC optimize(2) #include<cstdio> #include<iostream> #include<string> #include<cstring> #include<map> #include<cmath> #include<cctype> #include<vector> #include<set> #include<queue> #include<algorithm> #include<sstream> #include<ctime> #include<cstdlib> #define X first #define Y second #define L (u<<1) #define R (u<<1|1) #define pb push_back #define mk make_pair #define Mid (tr[u].l+tr[u].r>>1) #define Len(u) (tr[u].r-tr[u].l+1) #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 pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f; const double eps=1e-6;string s,ps; vector<int>v[28]; struct Node {int l,r;LL cnt,lazy; }tr[N<<2];void pushdown(int u) {if(!tr[u].lazy) return;LL lazy=tr[u].lazy; tr[u].lazy=0;tr[L].lazy+=lazy; tr[R].lazy+=lazy;tr[L].cnt+=lazy; tr[R].cnt+=lazy; }void build(int u,int l,int r) {tr[u]={l,r,1,0};if(l==r) {tr[u].cnt=l;return;}build(L,l,Mid); build(R,Mid+1,r); } void change(int u,int l,int r,int c) {if(tr[u].l>=l&&tr[u].r<=r) {tr[u].cnt+=c;tr[u].lazy+=c;return;}pushdown(u);if(l<=Mid) change(L,l,r,c);if(r>Mid) change(R,l,r,c); }int query(int u,int l,int r) {if(tr[u].l>=l&&tr[u].r<=r) return tr[u].cnt;int ans=0;pushdown(u);if(l<=Mid) ans+=query(L,l,r);if(r>Mid) ans+=query(R,l,r);return ans; }LL check() {LL ans=0;build(1,1,ps.length());for(int i=0;i<s.length();i++) {int pos=pos=v[s[i]-'a'].back(); v[s[i]-'a'].pop_back();LL now=query(1,pos,pos);ans+=abs(now-(i+1));change(1,1,pos,1);}return ans; }int main() {ios::sync_with_stdio(false);cin.tie(0);int _=1;while(_--) {int n; cin>>n;cin>>ps; s=ps; reverse(ps.begin(),ps.end());for(int i=ps.length()-1;i>=0;i--) {v[ps[i]-'a'].pb(i+1);}printf("%lld\n",check());}return 0; } /**/

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