codeforces869c The Intriguing Obsession

最新推荐文章于 2022-06-23 22:26:45 发布

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最新推荐文章于 2022-06-23 22:26:45 发布

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分类专栏： dp 暴力文章标签： dp 数学 codeforces 869c OI

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本文探讨了一个有趣的问题：在三个不同颜色的岛屿群中，如何计算符合特定条件的桥梁搭建方式的数量。通过动态规划的方法，文章提供了一种有效的解决方案。

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— This is not playing but duty as allies of justice, Nii-chan!

— Not allies but justice itself, Onii-chan!

With hands joined, go everywhere at a speed faster than our thoughts! This time, the Fire Sisters — Karen and Tsukihi — is heading for somewhere they've never reached — water-surrounded islands!

There are three clusters of islands, conveniently coloured red, blue and purple. The clusters consist of a, b and c distinct islands respectively.

Bridges have been built between some (possibly all or none) of the islands. A bridge bidirectionally connects two different islands and has length 1. For any two islands of the same colour, either they shouldn't be reached from each other through bridges, or the shortest distance between them is at least 3, apparently in order to prevent oddities from spreading quickly inside a cluster.

The Fire Sisters are ready for the unknown, but they'd also like to test your courage. And you're here to figure out the number of different ways to build all bridges under the constraints, and give the answer modulo 998 244 353. Two ways are considered different if a pair of islands exist, such that there's a bridge between them in one of them, but not in the other.

Input

The first and only line of input contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 5 000) — the number of islands in the red, blue and purple clusters, respectively.

Output

Output one line containing an integer — the number of different ways to build bridges, modulo 998 244 353.

   Examples 
 
     input 
    
      Copy 
    
1 1 1

     output 
   
8

     input 
    
      Copy 
    
1 2 2

     output 
   
63

     input 
    
      Copy 
    
1 3 5

     output 
   
3264

     input 
    
      Copy 
    
6 2 9

     output 
   
813023575

Note

In the first example, there are 3 bridges that can possibly be built, and no setup of bridges violates the restrictions. Thus the answer is 23 = 8.

In the second example, the upper two structures in the figure below are instances of valid ones, while the lower two are invalid due to the blue and purple clusters, respectively.

$\text{[math]}$ 题意：给你三个集合，每个集合内有点，规定统一集合内两点最短路要么为3要么不连通，求方案数。题解：我们定义dp[i][j]表示集合1加了i个点，集合2加了j个点时的方案数。显然 dp[i][j]=(dp[i-1][j]+(dp[i-1][j-1]*j)); 情况

#include <stdio.h>
long long dp[5001][5001],a,b,c;
int main()
{
    scanf("%I64d%I64d%I64d",&a,&b,&c);
    for(int i=0;i<=5000;i++) dp[i][0]=dp[0][i]=1;
    for(int i=1;i<=5000;i++)
        for(int j=1;j<=5000;j++)
            dp[i][j]=(dp[i-1][j]+(dp[i-1][j-1]*j)%998244353)%998244353;
    printf("%I64d\n",(dp[a][b]*dp[a][c])%998244353*dp[b][c]%998244353);
}

一种是加边，随意一个其他点，或者不加。 n^2可以通过本题数据。之后直接算答案就行了。