NOIP2014提高组.飞扬的小鸟

题目

512. 飞扬的小鸟

思路

不难发现状态可以表示为$f[i][j]$也就是到达$(i,j)$位置的最小点击数, 当前状态可以由上一个位置的状态转移, $f[i-1][j+y]$, 或者$f[i-1][j-k\cdot x]+k$转移, 计算时间复杂度$10000\times 1000\times 1000=10^{10}$必定超时, 因此需要将状态转移方程做等价变形

原状态转移方程如下
$$f[i][j]=\min (f[i-1][j+y],f[i-1][j-k\cdot x]+k)$$

将第二项展开得到

$$f[i-1][j-x]+1,f[i-1][j-2x]+2,...,f[i-1][j-kx]+k$$

再展开$j=j-x$的方程

$$f[i][j-x]=\min (f[i-1][j+y-x],f[i-1][j-2x]+1,f[i-1][j-3x]+2,...,f[i-1][j-kx]+k)$$

注意到从第二项开始的部分和$f[i][j]$的第二项开始的部分类似, 可以记为$g[i][j]$, 那么$g[i][j]=\min (g[i][j-x]+1,f[i-1][j-x]+1)$, 状态转移方程转化为$f[i][j]=\min (f[i-1][j+y],g[i][j])$, 空间换时间优化掉一维, 时间复杂度变为$10^7$可以通过

原始版本代码

#include 
#include 
#include 

using namespace std;

const int N = 10010, M = 1010, INF = 0x3f3f3f3f;

int n, m, cnt;
int x[N], y[N];
int pos[N], d[N], u[N];
int f[N][M], g[N][M];

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	cin >> n >> m >> cnt;
	for (int i = 1; i <= n; ++i) cin >> x[i] >> y[i];
	for (int i = 1; i <= cnt; ++i) {
		int pos, a, b;
		cin >> pos >> a >> b;
		d[pos] = a, u[pos] = b;
	}

	memset(f, 0x3f, sizeof f);
	memset(g, 0x3f, sizeof g);
	for (int i = 1; i <= m; ++i) f[0][i] = 0;

	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			if (j > x[i]) g[i][j] = min(g[i][j - x[i]] + 1, f[i - 1][j - x[i]] + 1);
			if (!u[i] || j > d[i] && j < u[i]) {
				f[i][j] = g[i][j];
				if (j + y[i] <= m) f[i][j] = min(f[i][j], f[i - 1][j + y[i]]);
			}
		}

		if (!u[i]) {
			f[i][m] = min(f[i][m], f[i - 1][m] + 1);
			for (int j = 1; j < m; ++j) {
				f[i][m] = min(f[i][m], f[i - 1][j] + (m - j + x[i] - 1) / x[i]);
			}
		}
	}

	int res = INF;
	for (int i = 1; i <= m; ++i) res = min(res, f[n][i]);

	if (res < INF >> 1) {
		cout << 1 << "\n";
		cout << res << "\n";
		return 0;
	}

	cout << 0 << "\n";
	for (int i = n; i >= 0; --i) {
		for (int j = 1; j <= m; ++j) {
			if (f[i][j] < INF >> 1) {
				int ans = 0;
				for (int k = 0; k <= i; ++k) {
					if (u[k]) ans++;
				}
				cout << ans << "\n";
				return 0;
			}
		}
	}

	return 0;
}

优化版本代码

#include 
#include 
#include 

using namespace std;

const int N = 10010, M = 1010, INF = 0x3f3f3f3f;

int n, m, cnt;
int x[N], y[N];
int pos[N], d[N], u[N];
int f[N][M], g[M];

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);

	cin >> n >> m >> cnt;
	for (int i = 1; i <= n; ++i) cin >> x[i] >> y[i];
	for (int i = 1; i <= cnt; ++i) {
		int pos, a, b;
		cin >> pos >> a >> b;
		d[pos] = a, u[pos] = b;
	}

	memset(f, 0x3f, sizeof f);
	for (int i = 1; i <= m; ++i) f[0][i] = 0;

	for (int i = 1; i <= n; ++i) {
		memset(g, 0x3f, sizeof g);
		for (int j = 1; j <= m; ++j) {
			if (j > x[i]) g[j] = min(g[j - x[i]] + 1, f[i - 1][j - x[i]] + 1);
			if (!u[i] || j > d[i] && j < u[i]) {
				f[i][j] = g[j];
				if (j + y[i] <= m) f[i][j] = min(f[i][j], f[i - 1][j + y[i]]);
			}
		}

		if (!u[i]) {
			f[i][m] = min(f[i][m], f[i - 1][m] + 1);
			for (int j = 1; j < m; ++j) {
				f[i][m] = min(f[i][m], f[i - 1][j] + (m - j + x[i] - 1) / x[i]);
			}
		}
	}

	int res = INF;
	for (int i = 1; i <= m; ++i) res = min(res, f[n][i]);

	if (res < INF >> 1) {
		cout << 1 << "\n";
		cout << res << "\n";
		return 0;
	}

	cout << 0 << "\n";
	for (int i = n; i >= 0; --i) {
		for (int j = 1; j <= m; ++j) {
			if (f[i][j] < INF >> 1) {
				int ans = 0;
				for (int k = 0; k <= i; ++k) {
					if (u[k]) ans++;
				}
				cout << ans << "\n";
				return 0;
			}
		}
	}

	return 0;
}