gym101667 2017-2018 ACM-ICPC, Asia Daejeon Regional Contest 简单题解

A:不会

B:暴力搜索,枚举每一次操作的情况,稍微剪枝,避免重复记录即可

#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn = 7;
int pz[maxn][maxn];
int lst[maxn];
ll res;
int ex,ey;
unordered_set<ll> win;
int check() {
  int f1 = 0, f2 = 0;
  for(int i = 1; i <= 4; i++) {
    for(int j = 1; j <= 2; j++) {
      if(pz[i][j] == 1 && pz[i][j+1] == 1 && pz[i][j+2] == 1)  f1++;
      if(pz[i][j] == 2 && pz[i][j+1] == 2 && pz[i][j+2] == 2)  f2++;
    }
  }
  for(int j = 1; j <= 4; j++)
  for(int i = 1; i <= 2; i++) {
    if(pz[i][j] == 1 && pz[i+1][j] == 1 && pz[i+2][j] == 1)  f1++;
    if(pz[i][j] == 2 && pz[i+1][j] == 2 && pz[i+2][j] == 2)  f2++;
  }
  for(int i = 1; i <= 2; i++) {
    for(int j = 1; j <= 2; j++) {
      if(pz[i][j] == 1 && pz[i+1][j+1] == 1 && pz[i+2][j+2] == 1)  f1++;
      if(pz[i][j] == 2 && pz[i+1][j+1] == 2 && pz[i+2][j+2] == 2)  f2++;
    }
  }
  for(int i = 1; i <= 2; i++) {
    for(int j = 3; j <= 4; j++) {
      if(pz[i][j] == 1 && pz[i+1][j-1] == 1 && pz[i+2][j-2] == 1)  f1++;
      if(pz[i][j] == 2 && pz[i+1][j-1] == 2 && pz[i+2][j-2] == 2)  f2++;
    }
  }
  if(f1 != 0)  return 1;
  return f2 == 0 ? 0 : 2;
}
int all;
void dfs(int p,int x,int y){
  if(pz[ex][ey]==1) return ;
  int tmp=check();
  if(tmp==1) return ;
  if(pz[ex][ey]==2&&p==0&&ex==x&&ey==y){
    if(tmp==2){
      ll st=3;
      rep(i,1,4)rep(j,1,4){
        st*=3;
        st+=pz[i][j];
      }
      win.insert(st);
      return ;
    }
  }
  if(tmp==2) return ;
  if(all==0) return ;
  rep(i,1,4){
    if(lst[i]==0) continue;
    lst[i]--;
    all--;
    int pos=1;
    while(pz[pos][i]!=0) pos++;
    pz[pos][i]=1+p;
    dfs(1-p,pos,i);
    pz[pos][i]=0;
    lst[i]++;
    all++;
  }
}
int ans[6][6][6];
int main() {
  memset(pz,0,sizeof pz);
  int a,b,c;cin>>a>>b>>c;
  rep(_i,1,4)lst[_i]=4;
  pz[1][a]=1;lst[a]=3;
  ex=b,ey=c;
  all=15;
  dfs(1,1,a);
  cout<<win.size();
}

 C:重定向边,记忆化搜索求DAG上最长路即可

    #include<bits/stdc++.h>
    #include <vector>
    #define ll long long
    #define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
    #define per(ii,a,b) for(int ii=b;ii>=a;--ii)
    #define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
    #define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mp make_pair
    #define pii pair<ll,ll>
    #define all(x) x.begin(),x.end()
    #define show(x) cout<<#x<<"="<<x<<endl
    #define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
    #define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
    #define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
    #define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
    #define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
    using namespace std;//head
    const int maxn=3e5+10,maxm=2e6+10;
    const ll INF=0x3f3f3f3f,mod=1e9+7;
    int casn,n,m,k;
    int d[maxn],dis[maxn],vis[maxn];
    pii e[maxn];
    int ans;
    vector<int> g[maxn];
    int dfs(int now){
      if(dis[now]) return dis[now];
      int res=1;
      for(int to:g[now]){
        res=max(res,dfs(to)+1);
      }
      return dis[now]=res;
    }
    int main(){
      cin>>n>>m;
      rep(i,1,m){
        int a,b;cin>>a>>b;
        e[i]=make_pair(a,b);
        d[a]++,d[b]++;
      }
      rep(i,1,m){
        int a=e[i].fi,b=e[i].se;
        if(d[a]==d[b]) continue;
        if(d[a]>d[b]) swap(a,b);
        g[b].push_back(a);
    //    vis[a]=1;
      }
      rep(i,1,n) if(!vis[i])dfs(i);
      rep(i,1,n) k=max(dis[i],k);
    //  showa(dis,1,n);
      cout<<k<<endl;
      return 0;
    }
     

 D:暴力

    #include<bits/stdc++.h>
    #include <vector>
    #define ll long long
    #define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
    #define per(ii,a,b) for(int ii=b;ii>=a;--ii)
    #define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
    #define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mp make_pair
    #define pii pair<ll,ll>
    #define all(x) x.begin(),x.end()
    #define show(x) cout<<#x<<"="<<x<<endl
    #define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
    #define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
    #define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
    #define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
    #define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
    using namespace std;//head
    const int maxn=1e3+10,maxm=2e6+10;
    const ll INF=0x3f3f3f3f,mod=1e9+7;
    int casn,n,m,k;
     
    bool vis[maxn];
     
    int f(ll x) {
      int res = 0;
      while(x) {
        int a = x%10;
        res += a*a;
        x /= 10;
      }
      return res;
    }
     
    int main(){
      ll n;
      cin >> n;
      if(n == 1) {
        cout << "HAPPY
";
        return 0;
      }
      memset(vis, false, sizeof vis);
      int now = f(n);
      vis[now] = true;
      while(now != 1) {
        now = f(now);
        if(vis[now]) {
          cout << "UNHAPPY
";
          return 0;
        }
        vis[now] = 1;
      }
      cout << "HAPPY
";
      return 0;
    }
     

 E:做法有点像枚举边求最小环

   枚举每一条边,把所有边权比它小的边放入图中,然后以这两个边为源点和汇点跑最小割即可

#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e6+10,maxm=2e6+10;
const ll INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k;
template<typename T>class mxf{public:
  struct node{int to,next;T cap;}e[maxm<<1];
  int cur[maxn],head[maxn],dis[maxn],gap[maxn];
  int nume=1,s,t,tot;
  void init(int n){
    rep(i,0,n) head[i]=gap[i]=dis[i]=0;
    nume=1;
  }
  void add(int a,int b,T c){
    e[++nume]={b,head[a],c};head[a]=nume;
    e[++nume]={a,head[b],0};head[b]=nume;
  }
  T dfs(int now,T flow=INF){
    if (now==t||!flow) return flow; 
    T use=0,tmp;
    int d=dis[now]-1,to;
    for (int &i=cur[now];i;i=e[i].next) {
      if(dis[to=e[i].to]==d&&(tmp=e[i].cap)){
        e[i].cap-=(tmp=dfs(to,min(flow-use,tmp)));
        e[i^1].cap+=tmp;
        if((use+=tmp)==flow) return use;
      }
    }
    if (!--gap[dis[now]]) dis[s]=tot+1; 
    ++gap[++dis[now]];
    cur[now]=head[now];
    return use;
  }
  T getflow(int ss,int tt,int n,T ans=0){
    tot=n;s=ss;t=tt;gap[0]=tot;
    memcpy(cur,head,(tot+1)<<2);
    while(dis[s]<=tot) ans+=dfs(s);
    return ans;
  }
};
mxf<int> net;
struct node{int a,b,c;}e[maxn];
bool cmp(node x,node y){
  return x.c<y.c;
}
int main() {
  cin>>n>>m;
  rep(i,1,m){
    cin>>e[i].a>>e[i].b>>e[i].c;
  }
  sort(e+1,e+1+m,cmp);
  ll ans=0;
  rep(i,1,m){
    net.init(n);
    int s=e[i].a,t=e[i].b;
    rep(j,1,m){
      if(j==i) continue;
      if(e[j].c>=e[i].c) break;
      net.add(e[j].a,e[j].b,1);
      net.add(e[j].b,e[j].a,1);
    }
    int tmp=net.getflow(s,t,n);
    ans+=tmp;
  }
  cout<<ans<<endl;
  return 0;
}
//	auto _start=chrono::high_resolution_clock::now();
//	auto _end=chrono::high_resolution_clock::now();
//  cerr<<"elapsed time: "<<chrono::duration<double,milli>(_end-_start).count()<<" ms
";

 F:递归分解,找规律

    #include<bits/stdc++.h>
    #include <vector>
    #define ll long long
    #define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
    #define per(ii,a,b) for(int ii=b;ii>=a;--ii)
    #define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
    #define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mp make_pair
    #define pii pair<ll,ll>
    #define all(x) x.begin(),x.end()
    #define show(x) cout<<#x<<"="<<x<<endl
    #define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
    #define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
    #define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
    #define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
    #define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
    using namespace std;//head
     
    pair<ll, ll> f(ll k, ll step) {
      ll len = (k), Size = len*len/4;
      if(k == 2) {
        if(step == 0)  return make_pair(1, 1);
        else if(step == 1)  return make_pair(1, 2);
        else if(step == 2)  return make_pair(2, 2);
        else return make_pair(2, 1);
      }
      int num = step/Size;
       step = step%Size;
      if(num == 0) {
        pair<ll, ll> tmp = f(k/2, step);
        swap(tmp.fi, tmp.se);
        return tmp;
      }
      else if(num == 1) {
        pair<ll, ll> tmp = f(k/2, step);
        tmp.se += len/2;
        return tmp;
      }
      else if(num == 2) {
        pair<ll, ll> tmp = f(k/2, step);
        tmp.fi += len/2;  tmp.se += len/2;
        return tmp;
      }
      else {
        pair<ll, ll> tmp = f(k/2, step);
        swap(tmp.fi, tmp.se);
        ll a = len/2-tmp.fi+1, b = len/2-tmp.se+1;
        return make_pair(a+len/2, b);
      }
    }
     
    int main() {
      std::ios::sync_with_stdio(false);
      ll k;  ll step;
      cin >> k >> step;
      step--;
      pair<ll, ll> t = f(k, step);
      cout << t.fi << " " << t.se << "
";
      return 0;
    }

G:不会

 H:一开始想了假算法,卡了很久,后来才写的fft

#include<bits/stdc++.h>
#define ll long long
#define rep(i, a, b) for(int i = a; i <= b; i++)
#define IO std::ios::sync_with_stdio(false)
using namespace std;
 
const int maxn=8e6+20,maxm=8e3+10;
const double pi=acos(-1.0);
struct cp{double x,y;}a[maxn],b[maxn];
cp operator*(cp a,cp b){return {a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x};}
cp operator+(cp a,cp b){return {a.x+b.x,a.y+b.y};}
cp operator-(cp a,cp b){return {a.x-b.x,a.y-b.y};}
class fourier{public:
  int rev[maxn],len,pw;
  void init(int n){
    len=1,pw=0;
    while(len<=n) len<<=1,pw++;
    rep(i,0,len-1) rev[i]=0;
    rep(i,0,len-1) rev[i]=(rev[i>>1]>>1)|((i&1)<<(pw-1));
  }
  void transform(cp*a,int flag){
    rep(i,0,len-1) if(i<rev[i]) swap(a[i],a[rev[i]]);
    for(int mid=1;mid<len;mid<<=1){
      cp wn={cos(pi/mid),flag*sin(pi/mid)};
      for(int r=mid<<1,j=0;j<len;j+=r){
        cp t={1,0};
        for(int k=0;k<mid;k++,t=t*wn){
          cp x=a[j+k],y=t*a[mid+j+k];
          a[j+k]=x+y,a[j+k+mid]=x-y;
        }
      }
    }
    if(flag==-1) rep(i,0,len-1) a[i].x/=len;
  }
}fft;
 
char st[] = {'R', 'S', 'P'};
int n, m;  string s, t;
int ans[maxn];
 
void solve(char ch, char ch1) {
   for(int i = 0; i <= fft.len; i++)
      a[i].x = b[i].x = a[i].y = b[i].y = 0;
   for(int i = 1; i <= m; i++) 
      if(t[m-i+1] == ch)  b[i].x = 1;
   for(int i = 1; i <= n; i++) 
      if(s[i] == ch1)  a[i].x = 1;
   fft.transform(a, 1);  fft.transform(b, 1);
   for(int i = 0; i < fft.len; i++) {
      a[i] = a[i]*b[i];
   }
   fft.transform(a, -1);
   for(int l = 0; l < n; l++) {
      ans[l] += (int)(a[l+m+1].x+0.5);
   }
}
 
int main() {
    std::ios::sync_with_stdio(false);
    cin >> n >> m >> s >> t;
    s = " " + s;  t = " " + t;
    fft.init(n+m);
    solve('R', 'S');  solve('S', 'P');  solve('P', 'R');
    int res = 0;
    for(int i = 0; i < n; i++)
      res = max(res, ans[i]);
    cout << res << "
";
    return 0;
}
 
/**
12 4
RSPPSSSRRPPR
RRRR
 
12 3
RRRRRRRRRRRR
SSS
 
12 4
PPPRRRRRRRRR
RSSS
 
12 4
RRRRRRRRRSSS
RRRS 
**/

I:KMP,枚举K即可

#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e6+10,maxm=2e6+10;
const ll INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k;
class prefix{public:
  int p[maxn],lens;
  int *s;
  void init(int *_s,int _lens){
    s=_s,lens=_lens;
    rep(i,0,lens-1) p[i]=0;
    p[0]=-1;
    int now=0,pos=-1;
    while(now<lens)
      if(pos==-1||s[now]==s[pos]) p[++now]=++pos;
      else pos=p[pos];
  }
  vector<int> find(int *t,int lent){
    int now,pos=0;
    vector<int> ans;
    while(now<lent) {
      if(pos==-1||t[now]==s[pos]) pos++,now++;
      else pos=p[pos];
      if(pos==lens) pos=p[pos],ans.push_back(now-lens);
    }
    return ans;
  }
}kmp;
int a[maxn];
int main() {IO;
  cin>>n;
  per(i,0,n-1){
    cin>>a[i];
  }
  kmp.init(a,n);
  int ans1=INF,ans2=INF;
  rep(i,1,n){
    int t=i-kmp.p[i];
    if(ans1+ans2==t+n-i){
      if(t<ans2) {
        ans1=n-i;ans2=t;
      }
    }else if(ans1+ans2>t+n-i){
      ans1=n-i;ans2=t;
    }
  }
  cout<<ans1<<' '<<ans2<<endl;
}

 J:不会

K:构造

#include<bits/stdc++.h>
#include <vector>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x,g,e) for(int i=g[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn = 1e4+10;
 
int a[maxn];
 
int main() {
  std::ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for(int i = 1; i <= n; i++) {
    int x;
    cin >> x >> a[i];
  }
  int nowx = 0, nowy = 0, l = 0, r = 0, u = 0, d = 0;
  int vx = 1, vy = 0;
  cout << "1 ";  r = 1;  nowx = 1;
  for(int i = 1; i < n; i++) {
    if(a[i] == 1) {
      if(vx == 1 && vy == 0) {
         vx = 0, vy = 1;
         cout << u+1-nowy << " ";
         nowy = u+1; u++;
      }
      else if(vx == 0 && vy == 1) {
         vx = -1, vy = 0;
         cout << nowx-(l-1) << " ";
         nowx = l-1;  l = l-1;
      }
      else if(vx == -1 && vy == 0) {
         vx = 0, vy = -1;
         cout << nowy-(d-1) << " ";
         nowy = d-1;  d--;
      }
      else {
         vx = 1, vy = 0;
         cout << r+1-nowx << " ";
         nowx = r+1;  r++;
      }
    }
    else if(a[i] == -1){
      if(vx == 1 && vy == 0) {
         vx = 0, vy = -1;
         cout << nowy-(d-1) << " ";
         nowy = d-1; d--;
      }
      else if(vx == 0 && vy == 1) {
         vx = 1, vy = 0;
         cout << r+1-nowx << " ";
         nowx = r+1;  r++;
      }
      else if(vx == -1 && vy == 0) {
         vx = 0, vy = 1;
         cout << u+1-nowy << " ";
         nowy = u+1;  u++;
      }
      else {
         vx = -1, vy = 0;
         cout << nowx-(l-1) << " ";
         nowx = l-1;  l--;
      }
    }
  }
}
 

 L:赛后补的,枚举天数进行dp,上限大概是n^3天

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 55;

ll dp[3][maxn][maxn*maxn*maxn], a[3][maxn];
vector<pair<int, int> > E[3][maxn];
int n[3], m[3], goal[3];

int main() {
    std::ios::sync_with_stdio(false);
    int p, all = 1;
    cin >> p;
    for(int k = 0; k < p; k++) {
        cin >> n[k] >> m[k];
        all *= n[k];
        for(int i = 1; i <= n[k]; i++)
            cin >> a[k][i];
        for(int i = 1; i <= m[k]; i++) {
            int x, y, c;
            cin >> x >> y >> c;
            //E[k][x].push_back(make_pair(y, c));
            E[k][y].push_back(make_pair(x, c)); //对每个点要知道谁能到他   反向存图
        }
        cin >> goal[k];
    }
    for(int i = 0; i < p; i++) {
        for(int j = 0; j <= n[i]; j++) {
            for(int k = 0; k <= all; k++) {
                dp[i][j][k] = 1e18;
            }
        }
    }
    #define fi first
    #define se second
    for(int k = 0; k < p; k++) {
     //   cout << "@@
";
        dp[k][1][0] = 0;
        for(int j = 1; j <= all; j++) {
            for(int i = 1; i <= n[k]; i++) {
                dp[k][i][j] = min(dp[k][i][j], dp[k][i][j-1]+a[k][i]);
                for(auto it : E[k][i]) {
                    int v = it.fi, c = it.se;
                    dp[k][i][j] = min(dp[k][i][j], dp[k][v][j-1]+c);
                }
            }
        }
    }
    ll ans = 1e18;
    for(int i = 1; i <= all; i++) {
        ll tmp = 0;
   // cout << "---i = " << i << endl;
        for(int k = 0; k < p; k++) {
    //        cout << "k = " << k << "    dp = " << dp[k][goal[k]][i] << endl;
            tmp += dp[k][goal[k]][i];
        }
        ans = min(ans, tmp);
    }
    cout << ans << "
";
}