poj1733

poj1733

题解：数据太大需要离散化，一道并查集的简单题

代码：

#include <vector>
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int const N = 10000 + 10;
int n,k,fa[N],sum[N],l[N],r[N],flag[N];
string s;
vector<int>ve;
void Init(){
	cin>>n>>k;
	ve.clear();
	for(int i=0;i<N;i++)
		fa[i] = i,sum[i] = 0;
	for(int i=1;i<=k;i++){
		cin>>l[i]>>r[i]>>s;
		--l[i];
		if(s[0] == 'e')	flag[i] = 0;
		else 	flag[i] = 1;
		ve.push_back(r[i]);
		ve.push_back(l[i]);
	}
	sort(ve.begin(),ve.end());
	ve.erase(unique(ve.begin(),ve.end()),ve.end());
}
int find(int x){
	if(x != fa[x]){
		int tmp = fa[x];
		fa[x] = find(fa[x]);
		sum[x] ^= sum[tmp];
	}
	return fa[x];
}
int main(){
	ios::sync_with_stdio(false);
	Init();
	int ans = 0;
	for(int i=1;i<=k;i++){
		int x = lower_bound(ve.begin(),ve.end(),l[i]) - ve.begin();
		int y = lower_bound(ve.begin(),ve.end(),r[i]) - ve.begin();
		int fx = find(x),	fy = find(y);
		if(fx != fy){
			fa[fy] = fx;
			sum[fy] = sum[x] ^ sum[y] ^ flag[i];
		}else{
			if(sum[x] ^ sum[y] != flag[i]){
				if(!ans){
					ans = i;
					break;
				}
			}
		}
	}
	if(!ans)	ans = k + 1;
	cout<<ans-1<<endl;
	return 0;
}