Time to Run

D - Time to Run

Codeforces Round #619 (Div. 2) Editorial

特别要考虑到只有一行或者只有一列的情况。

如果由 k 直接正向推路径的话，难免会要考虑到很多种情况，不如先得到全部走完的路径，然后再从末尾开始减，一直到 k。另外，可以对将要输出的向量进行修正，使得其中不会出现 f 为0的情况。

// Created by CAD on 2020/2/14.
#include <bits/stdc++.h>

#define fi first
#define se second
#define pis pair<int,string>
#define ll long long
using namespace std;

vector<pis> v;
void fix(){
    vector<pis> cop=v;
    v.clear();
    for(auto i:cop)
        if(i.fi)
            v.push_back(i);
}
int main()
{
    int n,m,k;  cin>>n>>m>>k;
    ll all=0;
    for(int i=1;i<=n;++i){
        v.push_back({m-1,"R"});
        all+=v.back().fi*v.back().se.size();
        if(i==1) v.push_back({m-1,"L"});
        else v.push_back({m-1,"UDL"});
        all+=v.back().fi*v.back().se.size();
        if(i==n) v.push_back({n-1,"U"});
        else v.push_back({1,"D"});
        all+=v.back().fi*v.back().se.size();
    }
    if(all<k) return puts("NO");
    else{
        puts("YES");
        while(all>k){
            string t=v.back().se;
            int cop=v.back().fi*v.back().se.size();
            v.pop_back();
            all-=cop;
            if(all>=k) continue;
            cop=k-all;
            v.push_back({cop/t.size(),t});
            if(cop%t.size())
            v.push_back({1,t.substr(0,cop%t.size())});
            all=k;
        }
        fix();
        cout<<v.size()<<'\n';
        for(auto i:v)
            cout<<i.fi<<" "<<i.se<<'\n';
    }
    return 0;
}

以及我超级暴力的写法....

// Created by CAD on 2020/2/14.
#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main()
{
    ll n,m,k;cin>>n>>m>>k;
    ll all=(4*n*m-2*n-2*m);
    if(all<k) return puts("NO");
    else {
        n--,m--;
        puts("YES");
        if((!n)||(!m)){
            ll t=n?n:m;
            if(k>t) cout<<2<<'\n';
            else cout<<1<<'\n';
            cout<<min(t,k)<<" "<<(t==n?'D':'R')<<"\n",k-=t;
            if(k>0) cout<<k<<" "<<(t==n?'U':'L')<<"\n";
            return 0;
        }
        ll x=n*(2*m+1),y=m*(2*n+1),laz=k-x-y,ans=0;
        if(k>x) {
            ans+=n*3;
            int cop=k;
            if(cop>x+y) cop-=laz;
            cop-=x;
            ans+=cop/(2*n+1)*3;
            cop%=(2*n+1);
            if(cop>0) ans++,cop--;
            if(cop>n) ans++;
            if(cop>0) ans++;
            if(laz>0) ans++;
            if(laz>m) ans++;
        }
        else{
            int cop=k;
            ans+=k/(2*m+1)*3;
            cop%=(2*m+1);
            if(cop>0) ans++;
            if(cop>m) ans++;
        }
        cout<<ans<<'\n';
        if(k<=x){
            ll cnt=k/(2*m+1);
            for(int i=1;i<=cnt;++i)
                cout<<m<<" R\n"<<m<<" L\n"<<"1 D\n";
            k=k%(2*m+1);
            int bj=0;
            while(k>0){
                if(bj==0)   cout<<min(k,m)<<" R\n",k-=m;
                else if(bj==1)  cout<<min(k,m)<<" L\n",k-=m;
                else cout<<"1 D\n",k--;
                bj++;
            }
        }
        else{
            if(k>x+y) k-=laz;
            for(int i=1;i<=n;++i)
                cout<<m<<" R\n"<<m<<" L\n"<<"1 D\n";
            k-=x;
            int cnt=k/(2*n+1);
            for(int i=1;i<=cnt;++i)
                cout<<"1 R\n"<<n<<" U\n"<<n<<" D\n";
            k=k%(2*n+1);
            int bj=2;
            while(k>0){
                if(bj==0)   cout<<min(k,n)<<" D\n",k-=n;
                else if(bj==1)  cout<<min(k,n)<<" U\n",k-=n;
                else cout<<"1 R\n",k--;
                bj--;
            }
            bj=0;
            while(laz>0){
                if(bj==0)   cout<<min(laz,m)<<" L\n",laz-=m;
                if(bj==1)   cout<<min(laz,n)<<" U\n",laz-=n;
                bj++;
            }
        }
    }
    return 0;
}