/*input
5 6
2 4 3 5
1 2 2 8
3 6 5 2
4 6 4 7
2 4 3 10
3 5
2 4 3 10
1 3 1 20
2 5 4 30
*/
#include <bits/stdc++.h>
using namespace std;
namespace my_template {
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;
#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;
template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }
template<typename T>
void print(vector<T> vec, string name = "") {
cout << name;
for (auto u : vec)
cout << u << ' ';
cout << '\n';
}
}
using namespace my_template;
const int MOD = 1000000007;
const ll INF = 1e14;
const int MX = 100101;
struct tral {
int a, b, c, d;
};
int main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int M, N;
cin >> M >> N;
vector<tral> sk(M);
for (int i = 0; i < M; ++i)
{
int a, b, c, d; cin >> a >> b >> c >> d;
sk[i] = {a, b, c, d};
}
reverse(sk.begin(), sk.end());
int l = sk[0].a;
int r = sk[0].b;
// for (int i = 1; i < M; ++i)
// {
// if (l <= sk[i].c and sk[i].c <= r) {
// l = min(l, sk[i].a);
// r = max(r, sk[i].b);
// }
// }
// printf("l = %d, r = %d\n", l, r);
// if (l != 1 or r != N) return !printf("-1\n");
ll ats = INF;
for (int m = 0; m < M; ++m)
{
ll suma = sk[m].d;
l = sk[m].a;
r = sk[m].b;
for (int i = m; i < M; ++i)
{
if (l <= sk[i].c and sk[i].c <= r and (sk[i].a < l or r < sk[i].b)) {
l = min(l, sk[i].a);
r = max(r, sk[i].b);
suma += sk[i].d;
}
}
if (l == 1 and r == N)
ats = min(ats, suma);
}
if (ats == INF) printf("-1\n");
else printf("%lld\n", ats);
}
/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Incorrect |
1 ms |
384 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Incorrect |
1 ms |
384 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Incorrect |
1 ms |
384 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Incorrect |
1 ms |
384 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
