[Segment tree Beats! || 分块] Codeforces 793F Tinkoff Challenge

我们用fi表示左边界是i的答案 一条线段[a,b]对答案的影响是 f1到fl中大于等于a的都跟b取max
这个可以用线段树科技做 复杂度证明跟区间最值操作的势能分析应该差不多

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cassert>using namespace std;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=100005;int T[N<<2],R[N<<2],F[N<<2],G[N<<2];inline void mark(int x,int a,int b){  if (T[x]>=a){    T[x]=b; F[x]=b; if (!G[x]) G[x]=a;  }}inline void calc(int &x,int &y,int &z){  if (z>x)    y=x,x=z;  else if (z<x && z>y)    y=z;}inline void upd(int x){  T[x]=R[x]=-1<<30;  calc(T[x],R[x],T[x<<1]);  calc(T[x],R[x],R[x<<1]);  calc(T[x],R[x],T[x<<1|1]);  calc(T[x],R[x],R[x<<1|1]);}inline void push(int x){  if (F[x]){    mark(x<<1,G[x],F[x]); mark(x<<1|1,G[x],F[x]);    F[x]=0; G[x]=0;  }}inline void Build(int x,int l,int r){  if (l==r){    T[x]=l; R[x]=-1<<30; return;  }  int mid=(l+r)>>1;  Build(x<<1,l,mid); Build(x<<1|1,mid+1,r);  upd(x);}inline void Modify(int x,int l,int r,int ql,int qr,int a,int b){  if (T[x]<a) return;  if (ql<=l && r<=qr && R[x]<a){    mark(x,a,b); return;  }  push(x);  int mid=(l+r)>>1;  if (ql<=mid) Modify(x<<1,l,mid,ql,qr,a,b);  if (qr>mid) Modify(x<<1|1,mid+1,r,ql,qr,a,b);  upd(x);}inline int Query(int x,int l,int r,int t){  if (l==r) return T[x];  push(x); int mid=(l+r)>>1;  if (t<=mid) return Query(x<<1,l,mid,t);  else return Query(x<<1|1,mid+1,r,t);}int n,m,pos[N];int ans[N];struct abcd{  int l,r,idx;  bool operator < (const abcd &B) const{    return r<B.r;  }}que[N];int main(){  int l,r;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n); read(m);  for (int i=1;i<=m;i++) read(l),read(r),pos[r]=l;  int Q; read(Q);  for (int i=1;i<=Q;i++){    read(que[i].l); read(que[i].r);    que[i].idx=i;  }  sort(que+1,que+Q+1);  int pnt=1;  Build(1,1,n);  for (int i=1;i<=n;i++){    if (pos[i])      Modify(1,1,n,1,pos[i],pos[i],i);    while (pnt<=Q && que[pnt].r==i)      ans[que[pnt].idx]=Query(1,1,n,que[pnt].l),pnt++;    //for (int j=1;j<=(n<<2);j++) assert(T[j]==0 || R[j]==0 || R[j]<T[j]);  }  for (int i=1;i<=Q;i++)    printf("%d\n",ans[i]);  return 0;}

标算给的是分块
里面有个单调栈理解上有点晦涩 具体我已经忘了 反正我知道我写的代码里那个栈里的东西似乎是不单调的 反而是另一个东西是单调的

#include<cstdio>#include<cstdlib>#include<algorithm>#include<vector>#define pb push_backusing namespace std;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=100005;const int BB=505;int n,B;vector<int> lr[N];int pos[N],lp[BB],rp[BB],tot;int f[BB][N],ans[BB][N];int Q[N],l,r;int pp[N];inline int calc(int l,int r,int t,int low){  for (int i=l;i<=r;i++)    if (pp[i] && pp[i]>=low)      if (t>=pp[i])    t=i;  return t;}int main(){  int n,m,l,r;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(n); read(m);  for (int i=1;i<=m;i++) read(l),read(r),lr[l].pb(r),pp[r]=l;  for (int i=1;i<=n;i++) sort(lr[i].begin(),lr[i].end());  B=sqrt(n);  for (int i=1;i<=n;i++) pos[i]=(i-1)/B+1; tot=pos[n];  for (int i=1;i<=tot;i++) lp[i]=(i-1)*B+1,rp[i]=i*B; rp[tot]=n;  for (int j=1;j<=n;j++){    int p=-1;    for (int i=pos[j];i<=tot;i++){      f[i][j]=j;      while (p+1<lr[j].size() && lr[j][p+1]<=rp[i]) p++;      if (~p) f[i][j]=lr[j][p];    }  }  for (int i=1;i<=tot;i++){    r=0;    for (int j=rp[i];j;j--){      ans[i][j]=f[i][j];      while (r && f[i][j]>=Q[r])    ans[i][j]=max(ans[i][j],ans[i][Q[r]]),r--;      Q[++r]=j;    }  }  int Q; read(Q);  while (Q--){    read(l); read(r);    if (pos[l]==pos[r]){      printf("%d\n",calc(l,r,l,l));      continue;    }    printf("%d\n",calc(lp[pos[r]],r,ans[pos[r]-1][l],l));  }  return 0;}