Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2)——B纯暴力—

本文分享了一次图论问题的解决经验，通过暴力求解而非搜索策略，实现了算法优化。作者通过对图的遍历，寻找最小的顶点度数之和，解决了特定的图论问题，展示了算法设计与优化过程。

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/*
写搜索WA了。。赛后发现直接暴力不会超啊，，，
*/
/************************************************
* Author        :Powatr
* Created Time  :2015-8-30 14:55:10
* File Name     :B.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 4e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;

int a[MAXN][MAXN];
vector<int> G[MAXN];
int main(){
    int n, m;
    while(~scanf("%d%d", &n, &m)){
        int x, y;
        memset(a, 0, sizeof(a));
        for(int i = 1; i <= n; i++)
            G[i].clear();
        for(int i = 1; i <= m; i++){
            scanf("%d%d", &x, &y);
            a[x][y] = a[y][x] = 1;
            G[x].push_back(y);
            G[y].push_back(x);
        }
        int min1 = INF;
        for(int i = 1; i <= n; i++){
            for(int j = i + 1; j <= n; j++){
                if(a[i][j]){
                    for(int k = j + 1; k <= n; k++){
                        if(a[i][k] && a[k][j]){
                    int temp = G[i].size() + G[j].size() + G[k].size();
                            min1 = min(min1,temp);
                        }
                    }
                }
            }
        }
        if(min1 == INF) puts("-1");
        else printf("%d\n", min1 - 6);
    }
    return 0;
}

转载于:https://www.cnblogs.com/zero-begin/p/4770907.html