CodeForces 402E Strictly Positive Matrix

本文介绍了一种将01矩阵问题转化为图论中强连通分量问题的解法,通过Tarjan算法判断有向图是否强连通,为解决特定类型的数学问题提供了新颖思路。

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题解:骚的一批。

首先一个很显然的问题是,把正数全部当成1,问题是等价的。所以这就变成了一个01矩阵。

重点来了,如果把这个01矩阵看成是一张有向图的邻接矩阵,那么这个矩阵的k次方,意思就是从i到j恰好经过k条边能否到达!

而这题又不要求你把k求出来,只问是否存在,所以只要整张图强连通,一定可以找到一个k满足这个条件(实在不行就把k设成两两点的路径长度的最小公倍数就行了)。于是问题转化成了给定一张有向图,问这张图是否强连通。用tarjan算法就可以完成了。

代码如下:

#include<bits/stdc++.h>
using namespace std;

int n,cnt,tcnt,R;
int dfn[2010],low[2010],S[2010];
bool f[2010][2010],ss[2010];

bool dfs(int x)
{
	dfn[x]=low[x]=++cnt;S[++R]=x;ss[x]=1;
	for(int i=1;i<=n;i++) if(f[x][i])
	{
	   if(!dfn[i]){dfs(i);low[x]=min(low[x],low[i]);}
	   else{if(ss[i]) low[x]=min(low[x],low[i]);}
	}
	if(dfn[x]==low[x])
	{
	  tcnt++;while(S[R]!=x){ss[S[R]]=0;R--;}
	  ss[x]=0;R--;
	}
}

int main()
{
	int a;
	scanf("%d",&n);
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++){scanf("%d",&a);f[i][j]=(a>0);}
	for(int i=1;i<=n;i++) if(!dfn[i]) dfs(i);
	if(tcnt>1) puts("NO");else puts("YES"); 
	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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