G. Rabbit Party

Time Limit: 5000ms

Case Time Limit: 5000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

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Rabbit Party

A rabbit Taro decided to hold a party and invite some friends as guests. He has n rabbit friends, and m pairs of
rabbits are also friends with each other. Friendliness of each pair is expressed with a positive integer. If two rabbits are not friends, their friendliness is assumed to be 0.

When a rabbit is invited to the party, his satisfaction score is defined as the minimal friendliness with any other guests. The satisfaction of the party itself is defined as the sum of satisfaction score for all the guests.

To maximize satisfaction scores for the party, who should Taro invite?

Write a program to calculate the maximal possible satisfaction score for the party.

Input

The first line of the input contains two integers, n and m (1
￥leq n ￥leq 100, 0 ￥leq m ￥leq 100). The rabbits are numbered from 1to n.

Each of the following m lines has three integers, u, v and f. u and v (1
￥leq u, v ￥leq n, u ￥neq v, 1 ￥leq f ￥leq 1,000,000) stands for the rabbits‘ number, and f stands
for their friendliness.

You may assume that the friendliness of a pair of rabbits will be given at most once.

Output

Output the maximal possible satisfaction score of the party in a line.

Sample Input 1

3 3
1 2 3
2 3 1
3 1 2

Output for the Sample Input 1

6

Sample Input 2

2 1
1 2 5

Output for the Sample Input 2

10

Sample Input 3

1 0

Output for the Sample Input 3

0

Sample Input 4

4 5
1 2 4
1 3 3
2 3 7
2 4 5
3 4 6

Output for the Sample Input 4

16

分析题目，因为给你的全然图（即一个顶点与其它的顶点都有边）

接着从题目能够分析的n * (n - 1) / 2 = m <= 100

（提示。假设一个点与其它点之间的友好值为0，那么能够去掉这个点，或者那个与它的边为0的点。大家能够思考一下，所以为零的能够直接去掉)

得知n <= 15所以，实际上仅仅有15个数，那么我们能够DFS，得到这些数都是不会超时的

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

#define pb push_back
#define mp make_pair
#define fillchar(a, x) memset(a, x, sizeof(a))
#define copy(a, b) memcpy(a, b, sizeof(a))
#define S_queue<P> priority_queue<P, vector<P>,greater<P> >

typedef long long LL;
typedef pair<int, int > PII;
typedef unsigned long long uLL;
template<typename T>
void print(T* p, T* q, string Gap = " "){int d = p < q ? 1 : -1;while(p != q){cout << *p;p += d;if(p != q) cout << Gap; }cout << endl;}
template<typename T>
void print(const T &a, string bes = "") {int len = bes.length();if(len >= 2)cout << bes[0] << a << bes[1] << endl;else cout << a << endl;}

const int INF = 0x3f3f3f3f;
const int MAXM = 1e2 + 5;
const int MAXN = 1e2 + 5;
int Fig[MAXN][MAXN], Max;
int X[25], n, m;

void Deal(int s){
    int f , ans = 0;
    for(int i = 1;i <= s;i ++){
        f = INF;
        for(int j = 1;j <= s;j ++){
            if(i == j) continue;
            f = min(f, Fig[X[i]][X[j]]);
        }
    if(f != INF)
        ans += f;
    }
    Max = max(ans, Max);
}
void DFS(int u){
    Deal(u);
    int st = X[u] + 1;
    if(u == 0) st = 1;
    for(int i = st;i <= n;i ++){
        bool flag = true;
        for(int j = 1;j < u;j ++){
            if(Fig[i][X[j]] == 0) {
                flag = false;
                break;
            }
        }
        X[u + 1] = i;
        if(flag) DFS(u + 1);
    }
}

int main(){
    int u, v, d;
    while(cin >> n >> m){
        Max = 0;
        memset(Fig, 0, sizeof(Fig));
        for(int i = 1;i <= m;i ++){
            cin >> u >> v >> d;
            Fig[u][v] = Fig[v][u] = d;
            Max = max(Max, d * 2);
        }
        DFS(0);
        print(Max);
    }
    return 0;
}