1301D - Time to Run（思维、模拟）

题目

思路：先进行一遍全部过程的模拟储存下来，为了方便模拟尽量先将相同的方向走完，而对于多少步step，将连续的相同方向当作一步，如DDDDDULLL->5D1U3L，只需判别存储好的全部模拟字符串前后不同步数+1即可，细节见代码。

#include<iostream>
#include<string>
#include<map>
#include<algorithm>
#include<memory.h>
#include<cmath>
#include<vector>
#define pii pair<int,int>
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const int Max = 1e6 + 5;
string ans[Max];
int s[Max];

int main()
{
    
    
	int n, m, k;cin >> n >> m >> k;
	if (k > n * m * 4 - 2 * n - 2 * m)cout << "NO" << endl;
	else
	{
    
    
		string str="";
		for (int i= 1;i <= n;i++)
		{
    
    
			for (int j = 1;j <= m - 1;j++)str += 'R';
			if (i != n)
			{
    
    
				for (int i = 1;i <= m - 1;i++)str += 'L';
				str += 'D';
			}
		}
		for (int i = 1;i <= m;i++)
		{
    
    
			for (int j = 1;j <= n - 1;j++)str += 'U';
			if (i != m)
			{
    
    
				for (int j = 1;j <= n - 1;j++)
				{
    
    
					str += 'D';
				}
				str += 'L';
			}
		 }
		int sum = 1;ans[sum] = str[0];s[sum] = 1;
		for (int i = 1;i <k;i++)
		{
    
    
			if (str[i] == str[i - 1])s[sum]++;
			else
			{
    
    
				ans[++sum] = str[i];
				s[sum] = 1;
			}
		}
		cout << "YES" << endl << sum << endl;
		for (int i = 1;i <= sum;i++)
		{
    
    
			cout << s[i] << " " << ans[i] << endl;
		}
	}
}