Codeforces 1181E1 A Story of One Country (Easy)

A Story of One Country (Easy)

考虑不断暴力用bfs地拆， 复杂度 n ^ 2 * log(n)， 1e5复杂度的不知道怎么做啊。。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n;

struct Node {
    int a, b, c, d;
    void read() {
        scanf("%d%d%d%d", &a, &c, &b, &d);
    }
    void print() {
        printf("%d %d %d %d ^^
", a, b, c, d);
    }
};

queue<vector<Node>> que;
vector<Node> now, tmpL, tmpR;

bool cmpa(const Node &x, const Node &y) {
    return x.a < y.a;
}
bool cmpb(const Node &x, const Node &y) {
    return x.c < y.c;
}

bool solve() {
    sort(ALL(now), cmpa);
    int maxB = now[0].b;
    for(int i = 1; i < SZ(now); i++) {
        if(now[i].a >= maxB) {
            tmpL.clear(); tmpR.clear();
            for(int j = 0; j < i; j++) tmpL.push_back(now[j]);
            for(int j = i; j < SZ(now); j++) tmpR.push_back(now[j]);
            que.push(tmpL); que.push(tmpR);
            return true;
        } else chkmax(maxB, now[i].b);
    }

    sort(ALL(now), cmpb);
    int maxD = now[0].d;
    for(int i = 1; i < SZ(now); i++) {
        if(now[i].c >= maxD) {
            tmpL.clear(); tmpR.clear();
            for(int j = 0; j < i; j++) tmpL.push_back(now[j]);
            for(int j = i; j < SZ(now); j++) tmpR.push_back(now[j]);
            que.push(tmpL); que.push(tmpR);
            return true;
        } else chkmax(maxD, now[i].d);
    }
    return false;
}

int main() {
    scanf("%d", &n);
    vector<Node> a(n);
    for(int i = 0; i < n; i++) a[i].read();
    que.push(a);
    while(!que.empty()) {
        now = que.front();
        que.pop();
        if(SZ(now) == 1) continue;
        if(!solve()) {
            puts("NO");
            return 0;
        }
    }
    puts("YES");
    return 0;
}

/*
*/