Codeforces Round#485

CF986A Fair（贪心+最短路）
CF986B Petr and Permutations（BIT+排列性质）

A

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 100010
inline char gc(){
    static char buf[1<<16],*S,*T;
    if(S==T){T=(S=buf)+fread(buf,1,1<<16,stdin);if(T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=gc();
    return x*f;
}
int n,m,K,S,h[N],num=0,dis[N];
struct edge{
    int to,next;
}data[N<<1];
priority_queue<int,vector<int>,greater<int> >qq[N];
vector<int>a[110];
inline void spfa(int id){
    queue<int>q;memset(dis,inf,sizeof(dis));
    for(int i=0;i<a[id].size();++i){
        int x=a[id][i];dis[x]=0;q.push(x);
    }while(!q.empty()){
        int x=q.front();q.pop();qq[x].push(dis[x]);
        for(int i=h[x];i;i=data[i].next){
            int y=data[i].to;if(dis[y]!=inf) continue;
            dis[y]=dis[x]+1;q.push(y);
        }
    }
}
int main(){
//  freopen("a.in","r",stdin);
    n=read();m=read();K=read();S=read();
    for(int i=1;i<=n;++i){
        int x=read();a[x].push_back(i);
    }while(m--){
        int x=read(),y=read();
        data[++num].to=y;data[num].next=h[x];h[x]=num;
        data[++num].to=x;data[num].next=h[y];h[y]=num;
    }
    for(int i=1;i<=K;++i) spfa(i);
    for(int i=1;i<=n;++i){
        int res=0;
        for(int ii=1;ii<=S;++ii){
            res+=qq[i].top();qq[i].pop();
        }printf("%d ",res);
    }return 0;
}

B

每对换一次，排列的奇偶性就会改变，可以算逆序对数或者n-轮换数来判断最终的排列的奇偶性，从而知道是3n还是7n+1.

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 1000010
inline char gc(){
    static char buf[1<<16],*S,*T;
    if(S==T){T=(S=buf)+fread(buf,1,1<<16,stdin);if(T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=gc();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=gc();
    return x*f;
}
int n,c[N];
ll res=0;
inline int ask(int x){
    int res=0;for(;x;x-=x&-x) res+=c[x];return res;
}
inline void add(int x){
    for(;x<=n;x+=x&-x) c[x]++;
}
int main(){
//  freopen("a.in","r",stdin);
    n=read();
    for(int i=1;i<=n;++i){
        int x=read();res+=ask(n-x+1);add(n-x+1);
    }puts((n&1)==(res&1)?"Petr":"Um_nik");
    return 0;
}