codeforces 339E Three Swaps (搜索 dfs)

题目链接:http://http://codeforces.com/problemset/problem/339/E


题意:

给出由1~n组成的序列,每次可将一个区间翻转

问如何从1~n的递增序列变成给出的序列,输出操作次数以及每次操作的区间

最多翻转3次,保证有解,输出的操作次数不一定是最小的


解题思路:

每次先翻转最长的连续区间


#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <math.h>
#include <algorithm>
using namespace std;
int a[1005];
int n,top=0;
struct node
{
	int li,ri;
}p[5];
bool check()
{
	for(int i=1;i<=n;i++)
	{
		if(a[i]!=i) return 0;
	}
	return 1;
}
bool dfs(int k)
{
	if(check()) 
	{
		printf("%d\n",top);
		for(int i=top;i>=1;i--)
		{
			printf("%d %d\n",p[i].li,p[i].ri);
		}
		return true; 
	}
	if(k==3) return false;
	for(int i=1;i<=n;i++)
	{
		int t1=abs(a[i]-a[i-1]);
		int t2=abs(a[i]-a[i+1]);
		if(a[i]!=i && (abs(a[i]-a[i-1])!=1 || abs(a[i]-a[i+1])!=1))
		{
			for(int j=i+1;j<=n;j++)
			{
				int t3=abs(a[j]-a[j-1]);
				int t4=abs(a[j]-a[j+1]);
				if(a[j]!=j && (abs(a[j]-a[j-1])!=1 || abs(a[j]-a[j+1])!=1))
				{
					top++;
					p[top].li=i;
					p[top].ri=j;
					reverse(a+i,a+j+1);
					if(dfs(k+1)) return true;
					reverse(a+p[top].li,a+1+p[top].ri);
					top--;
				}
			}
		}
	}
	return false;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
	}
	a[0]=a[n+1]=-1;
	top=0;
	dfs(0);
	return 0;
}



### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
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