Codeforces Round #439 (Div. 2) C The Intriguing Obsession (组合数学)

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本文解析了 Codeforces 平台上的题目 869C 的解题思路，通过理解题目中“至少3”的含义，将问题转换为确保不同颜色岛屿间最短距离不小于3的问题。文章进一步提出将图分为三部分来简化问题，并提供了一个 O(n^2) 的暴力求解方法，同时展示了如何将其优化到 O(n) 的效率。

tutiol: http://codeforces.com/blog/entry/55009

别人的代码;

Problem C

Consider islands of two colours and the bridges between them.
869C - The Intriguing Obsession

First step: Consider what does at least 3 mean?

'The shortest distance between them is at least 3' means it can't be 1 or 2. The distance can't be 1 means that no two islands with the same colour can be straightly connected. The distance can't be 2 means that for each island, no two islands with the same colour can both be straightly connected with it.

Second step: Make the graph into 3 parts.

The bridges between red and blue islands have no effection with those between red and purple ones. Therefore, we can make the graph into 3 parts: one between red and blue, one between blue and purple, and the last one between red and purple.

Suppuse there are A red islands and B blue islands, and there are k bridges between them. Then, the answer will be . So, the answer of bridges between red and blue ones should be

Therefore, the final answer should be ans1 * ans2 * ans3.

You can calculate it with an O(n2) brute force. Also, you can make it into O(n).

#include<cstdio>
#include<iostream>
using namespace std;
#define mod 998244353
#define maxn 5010
int C[maxn][maxn];
long long jie[maxn];
long long solve(int n,int m)
{
    long long ans=0;
    int lim=min(n,m);
    for(int i=0; i<=lim; i++)
        ans=(ans+1ll*C[n][i]*C[m][i]%mod*jie[i]%mod)%mod;
    return ans;
}
int main()
{
    C[0][0]=1;
    for(int i=1; i<=5000; i++)
    {
        C[i][0]=1;
        for(int j=1; j<=i; j++)
        {
            C[i][j]=(C[i-1][j]+C[i-1][j-1]);
            if(C[i][j]>=mod) C[i][j]-=mod;
        }
    }
    jie[0]=1;
    for(int i=1; i<=5000; i++)
        jie[i]=1ll*jie[i-1]*i%mod;
    int a,b,c; scanf("%d%d%d",&a,&b,&c);
    long long ans=solve(a,b)*solve(a,c)%mod;
    ans=ans*solve(b,c)%mod;
    cout<<ans<<endl;
    return 0;
}