[NOI2019]回家路线

题面

题解

同步赛时看到\(m \leq 2\times 10 ^ 5, q_i \leq 1000\)的我灵光一闪，交了个\(\mathrm{O}(mt)\)的大暴力然后\(\mathrm{AC}\)此题

设\(f[i][j]\)表示当前在第\(i\)个站点，时刻为\(j\)的最小烦躁度，\(F(x) = Ax^2 + Bx + C\)。

然后枚举每一条边\(i\)，可以得到：\(f[y_i][j] = \min\{f[x_i][p_i] + F(j - q_i)\}\)。

直接\(\mathrm{DP}\)即可。

代码

#include <cstdio>
#include <algorithm>

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if (ch == '-') w = -1, ch = getchar();
	while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int maxn(1e5 + 10), maxm(2e5 + 10), T(1010), INF(0x3f3f3f3f);
int n, m, A, B, C, maxQ, f[maxn][T];
struct node { int x, y, p, q; } L[maxm];
inline int operator < (const node &lhs, const node &rhs) { return lhs.q < rhs.q; }
inline int F(int x) { return x * x * A + x * B + C; }

void solve()
{
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= maxQ; j++) f[i][j] = INF;
	for (int j = 0; j <= maxQ; j++) f[1][j] = F(j);
	for (int i = 1; i <= m; i++)
	{
		int x = L[i].y;
		for (int j = L[i].q; j <= maxQ; j++)
			f[x][j] = std::min(f[x][j], f[L[i].x][L[i].p] + F(j - L[i].q));
	}
	int ans = INF;
	for (int j = 1; j <= maxQ; j++)
		ans = std::min(ans, f[n][j] + j - C);
	printf("%d\n", ans);
}

int main()
{
	n = read(), m = read(), A = read(), B = read(), C = read();
	for (int i = 1; i <= m; i++) L[i] = (node) {read(), read(), read(), read()};
	for (int i = 1; i <= m; i++) maxQ = std::max(maxQ, L[i].q);
	std::sort(L + 1, L + m + 1), solve();
	return 0;
}