版权声明：https://blog.csdn.net/qq_41730082 https://blog.csdn.net/qq_41730082/article/details/85094797
题目链接
判断哪一个最先出现的区间不符合要求的，就是我们知道有这样的区间even：偶数；odd：奇数。然后，就是用向量的方法向下推即可。
但是，这样会RE，一开始只看到了那个5000次的查询，所以当时就小了数组，然后发现是1e9的时候，就决定，离散化吧，然后就A了。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define efs 1e-7
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 2e4 + 7;
int N, Q, ans, L, R, root[maxN], sum[maxN], lsan[maxN], cnt = 0, lis_len;
char op[10];
struct node
{
int l, r, val;
node(int a=0, int b=0, int c=0):l(a), r(b), val(c) {}
}a[maxN];
bool cmp(int e1, int e2) { return e1<e2; }
int fid(int x)
{
if(x == root[x]) return x;
int tmp = root[x];
root[x] = fid(root[x]);
sum[x] = (sum[tmp] + sum[x] + 2)%2;
return root[x];
}
void init()
{
ans = Q; cnt = 0;
for(int i=0; i<maxN; i++) root[i] = i;
memset(sum, 0, sizeof(sum));
}
int main()
{
while(scanf("%d%d", &N, &Q)!=EOF)
{
init();
for(int i=1; i<=Q; i++)
{
scanf("%d%d%s", &L, &R, op);
L--;
a[i] = node(L, R, (op[0] == 'e'?0:1));
lsan[++cnt] = L;
lsan[++cnt] = R;
}
sort(lsan+1, lsan+1+cnt, cmp);
lis_len = (int)((unique(lsan+1, lsan+1+cnt) - lsan - 1));
for(int i=1; i<=Q; i++)
{
int tmp = a[i].val;
int l = (int)(lower_bound(lsan+1, lsan+1+lis_len, a[i].l) - lsan - 1), r = (int)(lower_bound(lsan+1, lsan+1+lis_len, a[i].r) - lsan - 1);
int u = fid(l), v = fid(r);
if(u != v)
{
root[v] = u;
sum[v] = (sum[l] - sum[r] + tmp + 2)%2;
}
else
{
if(((sum[r] - sum[l] + 2)%2!=tmp) && ans == Q) ans = i-1;
}
}
printf("%d\n", ans);
}
return 0;
}