思路：d（i, j）表示区间(i, j]的1的个数的奇偶性。输入最多共有5000*2个点，需要离散化处理一下。剩下的就是并查集判冲突。

AC代码

#include <cstdio>
#include <cmath>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 1e4 + 5;
int n, m;
struct node{
	int par;
	int real;
}a[maxn];

struct Edge{
	int u, v, k;
}b[maxn];

void init(int num) {
	for(int i = 0; i < num; ++i) {
		a[i].par = i;
		a[i].real = 0;
	}
}

int find(int x) {
	if(a[x].par == x) return x;
	int r = find(a[x].par);
	a[x].real = a[x].real ^ a[a[x].par].real;
	return a[x].par = r;
}

bool unionset(int x, int y, int real) {
	int rx = find(x), ry = find(y);
	if(rx == ry) {
		if(real != (a[x].real ^ a[y].real)) return false;
	}
	else {
		a[rx].par = y;
		a[rx].real = a[x].real ^ real;
	}
	return true;
}

void deal() { //离散化处理
	map<int, int>ha;
	vector<int>v;
	for(int i = 0; i < m; ++i) {
		if(!ha.count(b[i].u)) {
			ha[b[i].u] = 1;
			v.push_back(b[i].u);
		}
		if(!ha.count(b[i].v)) {
			ha[b[i].v] = 1;
			v.push_back(b[i].v);
		}
	}
	sort(v.begin(), v.end());
	int id = 0;
	for(int i = 0; i < v.size(); ++i) {
		ha[v[i]] = id++;
	}
	init(id);
	for(int i = 0; i < m; ++i) {
		b[i].u = ha[b[i].u];
		b[i].v = ha[b[i].v];
	}
} 

int main() {
	while(scanf("%d%d", &n, &m) == 2) {
		char s[10];
		for(int i = 0; i < m; ++i) {
			scanf("%d%d%s", &b[i].u, &b[i].v, s);
			b[i].u--;
			if(s[0] == 'o') b[i].k = 1; //奇数
			else b[i].k = 0;
		}
		deal();
		int ans = 0, flag = 1;
		for(int i = 0; i < m; ++i) {
			ans = i;
			if(!unionset(b[i].u, b[i].v, b[i].k)) {
				flag = 0;
				break;
			}
		}
		if(flag) ans = m;
		printf("%d\n", ans);
	}
	return 0;
} 

如有不当之处欢迎指出！