Title: POJ1733.
Given nnn intervals [l i, R i] [l_i, r_i] [l i, R i] and aia_iai, the weights of [l i, R i] [l_i, r_i] [l i, R i] and aia_iai are odd or even. Ask which interval is not contradictory, but its next interval is contradictory.
1≤n≤5∗1031\leq n\leq 5*10^31≤n≤5∗103.

It's easy to think of prefixes and sums, so intervals become information on two points li_1l_i-1li_1 and rir_iri.

Consider maintaining the prefix of each point and d[i]d[i]d[i], the point weight of a point I I I represents the prefix of I I I and the point weight of the father of exclusive or I I i, that is, the parity of the interval [fa[i]+1,i][fa[i]+1,i][fa[i]+1,i].

When an interval [l i, R i] [l_i, r_i] [l i, R i] is added, if l i_1, R I l_i-1, r_i L i_1 and R I are in the same interval, it will be vigorously judged whether their point weights and differences are equal to a[i]a[i]a[i] [i]. Then adding this interval will be equivalent to setting the ancestor of R I r of R I r_i R I as l i_1 l_i-1 and updating the point weights at the same time.

Note that at this time, the path compression also needs to update the point weight.

Time complexity O (n log n) O (n log n) O (n log n).

The code is as follows:

#include<iostream>
#include<cstdio>
#include<algorithm>
  using namespace std;

#define Abigail inline void
typedef long long LL;

const int N=5000;

int ri(){
  char c=getchar();
  int x=0,y=1;
  for (;c<'0'||c>'9';c=getchar()) if (c=='-') y=-1;
  for (;c<='9'&&c>='0';c=getchar()) x=x*10+c-'0';
  return x*y;
}

int rc(){
  char c=getchar();
  while (c<'a'||c>'z') c=getchar();
  return c=='o'?1:0;
}

int n,ord[N*2+9],x[N+9],y[N+9],a[N+9];
int fa[N*2+9],d[N*2+9],ans;

int lower(int k){
  int l=1,r=n<<1,mid=l+r>>1;
  for (;l<r;mid=l+r>>1)
    k<=ord[mid]?r=mid:l=mid+1;
  return l;
}

int get(int u){
  if (u==fa[u]) return u;
  int rot=get(fa[u]);d[u]^=d[fa[u]];
  return fa[u]=rot;
}

Abigail into(){
  ri();n=ri();
  for (int i=1;i<=n;++i){
  	x[i]=ri()-1;y[i]=ri();a[i]=rc();
  	ord[i*2-1]=x[i];ord[i*2]=y[i];
  }
}

Abigail work(){
  sort(ord+1,ord+1+2*n);
  for (int i=0;i<=n;++i)
    x[i]=lower(x[i]),y[i]=lower(y[i]);
  for (int i=1;i<=n<<1;++i) fa[i]=i;
  int u,v;
  ans=n;
  for (int i=1;i<=n;++i){
  	u=get(x[i]);v=get(y[i]);
  	if (u==v&&d[x[i]]^d[y[i]]^a[i]){ans=i-1;return;}
  	fa[v]=u,d[v]=d[x[i]]^d[y[i]]^a[i];
  }
}

Abigail outo(){
  printf("%d\n",ans);
}

int main(){
  into();
  work();
  outo();
  return 0;
}