洛谷 二分答案算法 P2678 [NOIP2015 提高组] 跳石头

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题目链接

链接: P2678 [NOIP2015 提高组] 跳石头

题目描述

解题思路

二分答案，就是二分距离，看是否满足移动次数少于m次并且可以让最短距离最大,详细思路见代码

代码实现

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 50010;
int L, n, m, a[N],b[N];
bool check(int x)
{
    //cout<<x<<endl;
	int cnt = 0;
	for (int i = 0; i <= (n+1); i++) b[i]=a[i];
	for (int i = 1; i <= (n+1); i++) {
		if (b[i] - b[i - 1] < x) {
			cnt++;
			b[i]=b[i-1];
			//cout << "挪动了第"<<i<<"块石头" << endl;
		}
	}
	if (cnt <= m) return true;
	else return false;
}
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin >> L >> n >> m;
	for (int i = 1; i <= n; i++) cin >> a[i];
	a[n + 1] = L;
	int l = 0, r = L;
	while (l < r)
	{
		int mid = (l + r + 1) >> 1;
		if (check(mid)) l = mid;
		else r = mid - 1;
	}
	cout << l;
	return 0;
}

总结

掌握二分答案算法