题目大意

题目链接

给你一个区间长度L (L <= 1e9) ， 和q(q <= 5000)组数据， 每组数据 x y odd/even ， 表示 区间[x, y]和为odd/even, 输出最先出现矛盾的组号(0 到 q - 1), 如果没有矛盾， 输出q

分析

如果 [x, y] 是偶数， 那么 [1, x) 和 [1, y] 同奇偶性， 即[1, x - 1] [1, y] 同奇偶性 即 x - 1 和 y是同类。

L太大， 需要离散化

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
/*
power by Solo_Dance
*/

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <vector>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>

#define eps 1e-8
using namespace std;
#define ms(a, b) memset((a), (b), sizeof(a))

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
const int N = 1e5 + 5;
const int M = 1e6 + 5;
const int INF = 0x3f3f3f3f;
const ll ll_max = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;

inline ll read() {
    ll res = 0;bool f = 0;char ch = getchar();
    while (ch < '0' || ch > '9') {if (ch == '-') f = 1;ch = getchar();}
    while (ch <= '9' && ch >= '0') {res = (res << 3) + (res << 1) + ch - '0';ch = getchar();}
    return f ? (~res + 1) : res;
}
namespace DSU{
    int f[N], siz[N];
    inline void init(){
        for (int i = 0; i < N; ++i) f[i] = i, siz[i] = 1;
    }
    inline int tofind(int x){
        if (f[x] != x) f[x] = tofind(f[x]);
        return f[x];
    }
    inline void tojoin(int a, int b){
        a = tofind(a), b = tofind(b);
        if (siz[a] > siz[b]) swap(a, b);
        f[a] = b; siz[b] = max(siz[b], siz[a] + 1);
    }
};
using namespace DSU;
vector<int>ve;
map<int, int>ma;
struct node{
    int l, r;
    string s;
}a[N];
int main(){
    int n = read();
    int q = read();
    for (int i = 1; i <= q; ++i){
        a[i].l = read() - 1, a[i].r = read();
        ve.push_back(a[i].l), ve.push_back(a[i].r);
        cin >> a[i].s;
    }
    
    // 离散化
    sort(ve.begin(), ve.end());
    ve.erase(unique(ve.begin(), ve.end()), ve.end());
    int sz = ve.size();
    for (int i = 0; i < sz; ++i) ma[ve[i]] = i + 1;
    
    int flag = 1;
    init();
    n = sz;
    for (int i = 1; i <= q; ++i){
        int x = ma[a[i].l], y = ma[a[i].r];
        if (a[i].s[0] == 'e'){
            if (tofind(x) == tofind(y + n)){
                cout << i - 1 << "\n";
                flag = 0;
                break;
            }
            tojoin(x, y), tojoin(x + n, y + n);
        }
        else{
            if (tofind(x) == tofind(y)){
                cout << i - 1 << "\n";
                flag = 0;
                break;
            }
            tojoin(x, y + n), tojoin(x + n, y);
        }
    }
    if (flag) cout << q << "\n";
    return 0;
}
恰似你一低头的温柔，较弱水莲花不胜寒风的娇羞， 我的心为你悸动不休。  --mingfuyan

千万不要图快——如果没有足够的时间用来实践， 那么学得快， 忘得也快。