G. Rabbit Party

Time Limit: 5000ms

Memory Limit: 65536KB

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

Submit Status PID: 39423

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 1 ton.

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 31 2 32 3 13 1 2

Output for the Sample Input 1

6

Sample Input 2

2 11 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 51 2 41 3 32 3 72 4 53 4 6

Output for the Sample Input 4

16

Submit Status PID: 39423

分析复杂度之后，暴搜满足条件的团即可。注意搜索的技巧与顺序，避免重复搜索。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define N 105#define M 105int G[N][N];int vis[N];int n, m;int ans;int S[N];void dfs(int u, int cnt){    if(cnt >= 2)    {        int sum = 0;        for(int i = 1; i <= cnt; i++)        {            int tmp = INF;            for(int j = 1; j <= cnt; j++)                if(i != j)                {                    tmp = min(tmp, G[S[i]][S[j]]);                }            sum += tmp;        }        ans = max(ans, sum);    }    for(int i = u + 1; i <= n; i++)    {        int ok = 1;        for(int j = 1; j <= cnt; j++)            if(!G[i][S[j]])            {                ok = 0;                break;            }        if(ok)        {            S[cnt + 1] = i;            dfs(i, cnt + 1);        }    }}int main(){    while(~scanf("%d%d", &n, &m))    {        memset(vis, 0, sizeof vis);        memset(G, 0, sizeof G);        int u, v, f;        for(int i = 0; i < m; i++)        {            scanf("%d%d%d", &u, &v, &f);            G[u][v] = G[v][u] = f;        }        ans = 0;        dfs(0, 0);        printf("%d\n", ans);    }    return 0;}/*3 31 2 32 3 13 1 22 11 2 54 51 2 41 3 32 3 72 4 53 4 61 0*/