Gym 100096D Guessing game
题面
Problem Description
Byteman is playing a following game with Bitman. Bitman writes down some 1 000 000 000-element
sequence of zeros and ones. Byteman’s task is to guess the sequence that Bitman created. He can achieve
this goal by asking Bitman questions of the form: ’What is the parity of the sum of a subsequence of your
sequence, that begins at the b-th element and ends at the e-th element of the sequence?’. After playing
the game for some time, Byteman started suspecting that Bitman was cheating. He would like to know
at which moment did Bitman first answer his question in an inconsistent way, so he asked you for help.
Write a program which:
-
reads a description of Byteman’s questions and Bitman’s answers,
-
computes the greatest number N, for which Bitman’s answers for the first N questions are consistent.
Input
The first line of the input file contains one integer n (0 ≤ n ≤ 100 000), denoting the number of Byteman’s
questions. Each of the following n lines describes one Byteman’s question with corresponding Bitman’s
answer in the form of three positive integers b, e and s (1 ≤ b ≤ e ≤ 1 000 000 000, s ∈ {0, 1}), separated
by single spaces. b and e are the indices of the first and the last element of the subsequence in Byteman’s
question. s = 0 means that Bitman answered that the sum was even and s = 1 — that it was odd.
Output
The first and only line of the output file should contain one integer — the greatest value of N such that
there exists a sequence of zeros and ones that is consistent with Bitman’s answers to first N Byteman’s
questions.
题意
给出一些关系，从l到r的1的个数为奇数还是偶数。问从哪组开始是矛盾的。
考虑一个点不是奇数就是偶数，用一个前缀和的形式表示到i为止的前面所有数i的个数是奇数还是偶数。将一个点拆成两个点。
如果一个点的奇数代表和偶数代表相连说明改组关系与之前的肯定矛盾。
若v==1，则将l-1的奇数代表和r的偶数代表相连，l-1的偶数代表和r的奇数代表相连，表示[l,r]中有改变奇偶性。
若v==1，则将l-1的奇数代表和r的奇数代表相连，l-1的偶数代表和r的偶数代表相连，表示[l,r]中未改变奇偶性。
考虑到b,e很大，用map离散化。
代码
#include <bits/stdc++.h>
using namespace std;
int n;
int fa[1000010];
int l[1000010];
int r[1000010];
int v[1000010];
int cnt;
map<int,int> f;
int ans;
int find(int k)
{
return fa[k]==k?k:fa[k]=find(fa[k]);
}
int main()
{
freopen("game.in","r",stdin);
freopen("game.out","w",stdout);
cin>>n;
for (int i=1;i<=n;i++)
{
cin>>l[i]>>r[i]>>v[i];
l[i]--;
if (!f[l[i]]) f[l[i]]=++cnt;
if (!f[r[i]]) f[r[i]]=++cnt;
}
for (int i=1;i<=cnt+cnt;i++) fa[i]=i;
for (int i=1;i<=n;i++)
{
int x=f[l[i]],y=f[r[i]];
if (v[i])
{
if (find(x)==find(y)) break;
ans++;
if (find(y+cnt)!=find(x)) fa[fa[x]]=fa[y+cnt];
if (find(x+cnt)!=find(y)) fa[fa[y]]=fa[x+cnt];
}
else
{
if (find(x)==find(y+cnt)) break;
ans++;
if (find(y)!=find(x)) fa[fa[x]]=fa[y];
if (find(x+cnt)!=find(y+cnt)) fa[fa[y+cnt]]=fa[x+cnt];
}
}
cout<<ans<<endl;
return 0;
}