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\(\mathcal{Solution}\)

设 \(f_i\) 为到达 \(i\) 的答案，不能到达则为 \(inf\)。

设 \(g_i\) 为考虑完前面的操作时，单独使用当前操作来到达 \(i\) 的最小步数，不能到达则为 \(inf\)。

每次读进一个操作就把 \(g\) dp 一次，然后更新 \(f\)。

具体的：

1. 初始化 \(g_i=\left\{\begin{matrix}0,f_i\neq inf\\inf,f_i=inf \end{matrix}\right.\)

2. 若 \(g_i\neq y\)，则 \(g_{\left \lceil i+x \right \rceil }=\min(g_{\left \lceil i+x \right \rceil },g_i+1)\) 或 \(g_{\left \lceil i*x \right \rceil }=\min(g_{\left \lceil i*x \right \rceil },g_i+1)\)。

3. 根据 \(g\) 更新 \(f\)，若 \(g_i\neq inf\)，则对 \(f_i\) 更新。

时间复杂度 \(\mathcal{O}(nm)\)

\(\mathcal{Code}\)

//Code by do_while_true
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
namespace do_while_true {
	#define ld double
	#define ll long long
	#define re register
	#define pb push_back
	#define fir first
	#define sec second
	#define pp std::pair
	#define mp std::make_pair
	const ll mod = 1e9 + 7;
	template <typename T>
	inline T Max(T x, T y) { return x > y ? x : y; }
	template <typename T>
	inline T Min(T x, T y) { return x < y ? x : y; }
	template <typename T>
	inline T Abs(T x) {	return x < 0 ? -x : x; }
	template <typename T>
	inline T& read(T& r) {
		r = 0; bool w = 0; char ch = getchar();
		while(ch < ‘0‘ || ch > ‘9‘) w = ch == ‘-‘ ? 1 : 0, ch = getchar();
		while(ch >= ‘0‘ && ch <= ‘9‘) r = r * 10 + (ch ^ 48), ch = getchar();
		return r = w ? -r : r;
	}
	template <typename T>
	inline T qpow(T x, T y) {
		re T sumq = 1; x %= mod;
		while(y) {
			if(y&1) sumq = sumq * x % mod;
			x = x * x % mod;
			y >>= 1;
		}
		return sumq;
	}
	char outch[110];
	int outct;
	template <typename T>
	inline void print(T x) {
		do {
			outch[++outct] = x % 10 + ‘0‘;
			x /= 10;
		} while(x);
		while(outct >= 1) putchar(outch[outct--]);
	}
}
using namespace do_while_true;

const int N = 1e5 + 10;
const int INF = 0x3ffffff;

int n, m;
int f[N], g[N];

void work(int q) {
	int t, y; ll x;
	for(int i = 1; i <= m; ++i) g[i] = INF;
	read(t); read(x); read(y);
	if(t == 1) {
		x = x / 100000 + (x % 100000 > 0);
		g[x] = 1;
		for(int i = 1; i <= m; ++i)
			if(f[i] != INF) g[i] = 0;
		for(int i = 1; i <= m; ++i)
			if(g[i] != y && g[i] != INF && i+x <= m)
				g[i+x] = Min(g[i+x], g[i]+1);
		for(int i = 1; i <= m; ++i)
			if(g[i] <= y && f[i] == INF)
				f[i] = q;
	}
	else {
		for(int i = 1; i <= m; ++i)
			if(f[i] != INF) g[i] = 0;
		for(int i = 1; i <= m; ++i) {
			int p = (int)std::ceil(1.0 * i * x / 100000);
			if(g[i] != y && g[i] != INF && p <= m)
				g[p] = Min(g[p], g[i]+1);
		}
		for(int i = 1; i <= m; ++i)
			if(g[i] <= y && f[i] == INF)
				f[i] = q;
	}
}

signed main() {
	read(n); read(m);
	for(int i = 1; i <= m; ++i) f[i] = INF;
	for(int i = 1; i <= n; ++i)
		work(i);
	for(int i = 1; i <= m; ++i) printf("%d ", f[i] == INF ? -1 : f[i]);
	return 0;
}

Codeforces 1498D Bananas in a Microwave 题解

标签：==   class   ceil   second   pen   tchar   i+1   iostream   oid   

原文地址：https://www.cnblogs.com/do-while-true/p/14622348.html