Counting Cliques HDU - 5952

本文介绍了一种求解无向图中特定大小完全子图数量的算法。通过深度优先搜索(DFS)策略,结合剪枝技巧避免重复计算,有效找出所有符合条件的完全图分量。

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题目:https://vjudge.net/contest/194395#problem/E

题目大意

给你一个无向图,让你求其中大小为s的完全图分量。

分析

暴力来做,设置一个集合,dfs当前点是否与集合中的点都有边,边界为集合大小,会超时,需要剪枝
每次都选择节点数大的点进去,这样就不会重复之前的选择了,比如1,2这种情况选择过了,然后2,1你就不用考虑了。

训练的时候想的是先把图补边为完全图,然后将没有的边从全排列里面剔除出去,然后各种麻烦弄不出来,还带偏了队友的思路呵呵

代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <ctype.h>
#include <set>
#include <vector>
#include <cmath>
#include <bitset>
#include <algorithm>
#include <climits>
#include <string>
#include <list>
#include <cctype>
#include <cstdlib>
#include <fstream>
#include <sstream>
using namespace std;
#define ll long long
#define MAX 105

ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n, m ,s;
int res;
bool mmap[MAX][MAX];
vector<int> G[MAX];
//set<int> S;
//set<int>::iterator iter;
bool vis[MAX];

//bool check(int v) {
//    for(iter = S.begin(); iter != S.end(); iter++) {
//        if(!mmap[*iter][v]){
//            return false;
//        }
//    }
//    return true;
//}


void dfs(int a[],int u, int now) {
    if(now >= s) {
        res++;
        return;
    }

    for(int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if(!vis[v]&&a[now]<v) 
        {
            int flag=1;
            for(int i=1;i<=now;i++)
            {
                if(mmap[a[i]][v]!=1)
                {
                    flag=0;
                    break;
                }
            }
            if(flag)
            {
                a[now+1]=v;
                vis[v] = 1;
                dfs(a, v, now+1);
                vis[v] = 0;
            }
        }
    }
}


int main() {
    int T;
    //ios::sync_with_stdio(false);
    //cin >> T;
    scanf("%d",&T);
    while(T--) {
        //cin >> n >> m >> s;
        scanf("%d%d%d",&n,&m,&s);
        //MEM(mmap, 0);
        memset(mmap,0,sizeof(mmap));
        for(int i = 1; i <= n; i++) {
            vis[i] = 0;
            G[i].clear();
        }
        res = 0;

        for(int i = 0; i < m; i++) {
            int a, b;
            //cin >> a >> b;
            scanf("%d%d",&a,&b);
            mmap[a][b] = 1;
            mmap[b][a] = 1;
            G[a].push_back(b);

        }
        int shu[105];
        for(int i = 1; i <= n; i++){
            shu[1]=i;
            dfs(shu,i,1);
        }
        //cout << res << endl;
        printf("%d\n",res);
    }
}
### HDU 1466 Problem Description and Solution The problem **HDU 1466** involves calculating the expected number of steps to reach a certain state under specific conditions. The key elements include: #### Problem Statement Given an interactive scenario where operations are performed on numbers modulo \(998244353\), one must determine the expected number of steps required to achieve a particular outcome. For this type of problem, dynamic programming (DP) is often employed as it allows breaking down complex problems into simpler subproblems that can be solved iteratively or recursively with memoization techniques[^1]. In more detail, consider the constraints provided by similar problems such as those found in references like HDU 6327 which deals with random sequences using DP within given bounds \((1 \leq T \leq 10, 4 \leq n \leq 100)\)[^2]. These types of constraints suggest iterative approaches over small ranges might work efficiently here too. Additionally, when dealing with large inputs up to \(2 \times 10^7\) as seen in reference materials related to counting algorithms [^4], efficient data structures and optimization strategies become crucial for performance reasons. However, directly applying these methods requires understanding how they fit specifically into solving the expectation value calculation involved in HDU 1466. For instance, if each step has multiple outcomes weighted differently based on probabilities, then summing products of probability times cost across all possible states until convergence gives us our answer. To implement this approach effectively: ```python MOD = 998244353 def solve_expectation(n): dp = [0] * (n + 1) # Base case initialization depending upon problem specifics for i in range(1, n + 1): total_prob = 0 # Calculate transition probabilities from previous states for j in transitions_from(i): # Placeholder function representing valid moves prob = calculate_probability(j) next_state_cost = get_next_state_cost(j) dp[i] += prob * (next_state_cost + dp[j]) % MOD total_prob += prob dp[i] %= MOD # Normalize current state's expectation due to accumulated probability mass if total_prob != 0: dp[i] *= pow(total_prob, MOD - 2, MOD) dp[i] %= MOD return dp[n] # Example usage would depend heavily on exact rules governing transitions between states. ``` This code snippet outlines a generic framework tailored towards computing expectations via dynamic programming while adhering strictly to modular arithmetic requirements specified by the contest question format.
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