gym101908 [2018-2019 ACM-ICPC Brazil Subregional Programming Contest] 题解

Posted 2020-11-13 nervendnig

感觉难度和邀请赛类似,题目质量也低于国内邀请赛,(题面/数据不出锅的情况下)

https://codeforces.com/gym/101908

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A.大概是莫比乌斯之类的,不会

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B:博弈,不会

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C:欧拉公式+二维偏序

首先,根据平面图欧拉公式,可推导出答案为$n+m+1+$交叉的数量

交叉的数量由三部分构成,横竖交叉数,横横交叉数和竖竖交叉数

显然,前者为$nm$,后两个为简单的二维偏序问题,

比较扯淡的是,先T了一发,从map改完二分,交的时候才意识到是没关同步

#include<bits/stdc++.h>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define fi first
#define se second
#define pii pair<int,int>
#define all(x) x.begin(),x.end()
using namespace std;//head
const int maxn=1e6+10,maxm=2e6+10;
const ll INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k;
ll x,y;
class bit{public:
	int node[maxn];
	inline int lb(int x) {return x&(-x);}
	void init(int n){fill_n(node,n+1,0);}
  inline void update(int pos,int val,int n){for(;pos>0&&pos<=n;pos+=lb(pos)) node[pos]+=val;}
  inline int ask(int pos,int sum,int n){for(;pos>0;pos-=lb(pos)) sum+=node[pos];return sum;}
  inline int query(int l,int r,int n){return ask(r,0,n)-ask(l-1,0,n);}
}tree;
int px[maxn];
ll solve(int n){
  int cntx=0;
  vector<pii>h;
  rep(i,1,n){
    int a,b;cin>>a>>b;
    h.emplace_back(a,b);
    px[++cntx]=b;
  }
  sort(px+1,px+1+cntx);
  cntx=unique(px+1,px+1+cntx)-px-1;
  for(auto &i:h){
    i.se=lower_bound(px+1,px+1+cntx,i.se)-px;
  }
  ll cnt=0;
  sort(all(h));
  tree.init(cntx);
  for(auto &i:h){
    cnt+=tree.query(i.se+1,cntx,cntx);
    tree.update(i.se,1,cntx);
  }
  return cnt;
}
int main() {IO;
  cin>>x>>y>>n>>m;
  ll ans=solve(n)+solve(m)+n+m+1+1ll*n*m;
  cout<<ans<<endl;
}

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D题:签到

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E题:签到(30ms跑1e8已经是cf的基本操作了吗?)

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F题:状压DP

先对于节目,按结束时间排序,然后依次枚举:节目数量,节目,选择的$Festival$状态,

复杂度$(2^n)(sum{M_i})^2$,大概是1e9,最后运行时间500ms,只能说CF评测机牛逼了

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 86500;
int dp[1100][1<<10];
struct node {int st, et, val, id;}p[1100]; 
bool cmp(node x, node y) {return x.et < y.et;}
int main() {
  std::ios::sync_with_stdio(false);
  int n,cnt = 0, ma = 0,ans = -1;
  cin >> n;
  for(int i = 0; i < n; i++) {
    int k;cin >> k;
    for(int j = 0; j < k; j++) {
      int a, b, c, d;
      cin >> a >> b >> c;  d = i;
      p[++cnt] = {a, b, c, d};
      ma = max(ma, b);
    }
  }
  sort(p+1, p+1+cnt, cmp);
  memset(dp, -1, sizeof dp);
  dp[0][0] = 0;
  for(int i = 1; i <= cnt; i++) {
    dp[i][0] = 0;
    for(int j = 0; j < i; j++) {
      if(p[i].st < p[j].et)  continue;
      for(int k = 1; k < (1<<n); k++) {
        if((k&(1<<p[i].id))) {
          if(dp[j][k] != -1)  dp[i][k] = max(dp[i][k], dp[j][k]+p[i].val);
          if(dp[j][k-(1<<p[i].id)] != -1)  dp[i][k] = max(dp[i][k], dp[j][k-(1<<p[i].id)]+p[i].val);
        }
      }
    }
    ans = max(ans, dp[i][(1<<n)-1]);
  }
  cout << ans << "
";
}

 赛后经过研究,可以优化为$(2^n)sum{M_i}$,实测31ms

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 86500;
int dp[1100][1<<10], ma[1100][1<<10],n, cnt;
struct node {int st, et, val, id;}p[1100];
bool cmp(node x, node y) {return x.et < y.et;}
int f(int R, int k) {
  int ans = 0, l = 1, r = R-1;
  while(l <= r) {
    int mid = (l+r)>>1;
    if(p[mid].et <= k)  ans = mid, l = mid+1;
    else r = mid-1;
  }
  return ans;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin >> n;
  for(int i = 0; i < n; i++) {
    int k;cin >> k;
    for(int j = 0; j < k; j++) {
      int a, b, c, d;
      cin >> a >> b >> c;  d = i;
      p[++cnt] = {a, b, c, d};
    }
  }
  sort(p+1, p+1+cnt, cmp);
  memset(dp, -1, sizeof dp);
  memset(ma, -1, sizeof ma);
  ma[0][0] = dp[0][0] = 0;
  int ans = -1;
  for(int i = 1; i <= cnt; i++) {
    dp[i][0] = 0;
    int sign = f(i, p[i].st);
    for(int k = 1; k < (1<<n); k++) {
      if(!(k&(1<<p[i].id)))  continue;
      if(ma[sign][k] != -1)  dp[i][k] = max(dp[i][k], ma[sign][k]+p[i].val);
      if(ma[sign][k-(1<<p[i].id)] != -1)  dp[i][k] = max(dp[i][k], ma[sign][k-(1<<p[i].id)]+p[i].val);
    }
    for(int k = 0; k < (1<<n); k++) ma[i][k] = max(ma[i-1][k], dp[i][k]);
  }
  cout << ma[cnt][(1<<n)-1] << "
";
  return 0;
}

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G:二分+最小割

裸题,显然答案为使用过的最大的边,那么二分答案,看最小割是否等于$sum{D_i}$即可,数组开小RE了一发

#include<bits/stdc++.h>
#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 IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;//head
const int maxn=1e5+10,maxm=2e6+10;
const int INF=0x3f3f3f3f,mod=1e9+7;
int casn,n,m,k,root;
class mxf{public:
  struct node{int to,next;int cap;}e[maxm<<1];
  int cur[maxn],head[maxn],que[maxn],dis[maxn],nume=1,s,t;
  inline void adde(int a,int b,int c){e[++nume]={b,head[a],c};head[a]=nume;}
  inline void add(int a,int b,int c){adde(a,b,c);adde(b,a,0);}
  void init(int n=maxn-1){memset(head,0,(n+1)<<2);nume=1;}
  bool bfs(){
  memset(dis,-1,(t+1)<<2);
    dis[t]=0,que[0]=t;
    int tp=0,ed=1;
    while(tp!=ed){
      int now=que[tp++];tp%=maxn;
      for(int i=head[now];i;i=e[i].next){
      int to=e[i].to;
      if(dis[to]==-1&&e[i^1].cap){
        dis[to]=dis[now]+1;
        if(to==s) return true;
        que[ed++]=to;ed%=maxn;
      }
      }
    }
    return false;
  }
  int dfs(int now,int flow=0x3f3f3f3f){
    if(now==t||flow==0) return flow;
    int use=0;
    for(int &i=head[now];i&&use!=flow;i=e[i].next){
      int to=e[i].to;
      if(dis[to]+1!=dis[now])continue;
      int tmp=dfs(to,min(e[i].cap,flow-use));
      e[i].cap-=tmp,e[i^1].cap+=tmp,use+=tmp;
    }
    if(!use) dis[now]=-1;
    return use;
  }
  int getflow(int ss,int tt){
    s=ss,t=tt;int ans=0;
    memcpy(cur,head,(t+1)<<2);
    while(bfs()){
      ans+=dfs(s);
      memcpy(head,cur,(t+1)<<2);
    }
    return ans;
  }
}net;
int need[maxn],have[maxn];
int sum,a[maxn],b[maxn],c[maxn];
bool check(int mid){
  net.init(n+m+10);
  int ss=n+m+1,tt=ss+1;
  rep(i,1,k)if(c[i]<=mid)net.add(a[i]+n,b[i],INF);
  rep(i,1,n)net.add(i,tt,need[i]);
  rep(i,1,m)net.add(ss,i+n,have[i]);
  return net.getflow(ss,tt)>=sum;
}
int main() {IO;
  cin>>n>>m>>k;
  rep(i,1,n)(cin>>need[i]),sum+=need[i];
  rep(i,1,m)cin>>have[i];
  rep(i,1,k)cin>>b[i]>>a[i]>>c[i];
  int l=1,r=1e6;
  int ans=-1;
  while(l<=r){
    int mid=(l+r)>>1;
    if(check(mid)) ans=mid,r=mid-1;
    else l=mid+1;
  }
  cout<<ans<<endl;
}

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H题:没看

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I题:签到

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J题:没看

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I题:计算几何,不会

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L题:树上链交裸题

一种做法是用树链剖分,单次查询是$log^2n$,很裸也很简单

但是这次我尝试写了单次$logn$直接讨论情况的做法,由于本题$Q$很小,拉不开差距,

具体情况大概分为3类,讨论了很多发才过

#include<bits/stdc++.h>
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define forn(i,x) for(int i=head[x];i;i=e[i].next)
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;//head
const int maxn=2e5+10,maxm=2e6+10;
int casn,n,m,k,root;
class graph{public:
  struct node{int to,next;}e[maxn<<1];
  int head[maxn],nume,dfn[maxn];
  inline void add(int a,int b){
    e[++nume]={b,head[a]};
    head[a]=nume;
  }
  int ltop[maxn],fa[maxn],deep[maxn];
  int sz[maxn],remp[maxn];
  int son[maxn],cnt;
  void init(int n){rep(i,1,n) head[i]=0;cnt=0,nume=1;}
  void dfs1(int now=root,int pre=root,int d=0){
    deep[now]=d,fa[now]=pre,sz[now]=1,son[now]=0;
    forn(i,now){
      int to=e[i].to;
      if(to!=pre) {
        dfs1(to,now,d+1);
        sz[now]+=sz[to];
        if(sz[to]>sz[son[now]]) son[now]=to;
      }
    }
  }
  void dfs2(int now=root,int pre=root,int sp=root){
    ltop[now]=sp;dfn[now]=++cnt;remp[cnt]=now;
      if(son[now])  dfs2(son[now],now,sp);
      forn(i,now){
        int to=e[i].to;
        if(to!=son[now]&&to!=pre) dfs2(to,now,to);
      }
  }
  void getchain(){dfs1();dfs2();}
  int lca(int x,int y){
    for(;ltop[x]!=ltop[y];deep[ltop[x]]>deep[ltop[y]]?x=fa[ltop[x]]:y=fa[ltop[y]]);
    return deep[x]<deep[y]?x:y;
  }
  inline int getdis(int a,int b){return deep[a]+deep[b]-2*deep[lca(a,b)];}
  bool check(int a,int b){return dfn[a]<=dfn[b]&&dfn[a]+sz[a]-1>=dfn[b]+sz[b]-1;}
  int getans(int a1,int a2,int b1,int b2){
    int ra=lca(a1,a2);
    bool f1=check(ra,b1),f2=check(ra,b2);
    if(!f1&&!f2) return 0;
    if(f1&&f2){
      int rb=lca(b1,b2);
      if(!( check(rb,a1)||check(rb,a2)))return 0;
      int r1=lca(a1,b1),r2=lca(a1,b2),r3=lca(a2,b1),r4=lca(a2,b2);
      if(r1==r3&&r2==r4) return 1;
      return getdis(r1==ra?r3:r1,r2==ra?r4:r2)+1;
    }
    if(!f1)swap(b1,b2);
    int r1=lca(a1,b1),r3=lca(a2,b1);
    return getdis(r1==ra?r3:r1,ra)+1;
  }
}g;
int d[maxn],a,b,a1,a2,b1,b2;
int main() {IO;
  cin>>n>>m;
  rep(i,2,n){
    cin>>a>>b;
    g.add(a,b);g.add(b,a);
    d[a]++,d[b]++;
  }
  rep(i,1,n) if(d[i]>1) {root=i;break;}
  g.getchain();
  while(m--){
    cin>>a1>>a2>>b1>>b2;
    int ans=g.getans(a1,a2,b1,b2);
    cout<<ans<<endl;
  }
}

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M题:没看

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