DFS 专题 Accepted Necklace

这道题如果用 DFS 必须剪枝：剪掉重复的组合。DFS 枚举了所有的排列，实际上这道题只需要组合。

View Code

# include <cstdio>

# include <cstring>



# define N 20 + 5



int ans;

int n, k, vMax, v[N], w[N];

bool vis[N];



void dfs(int cur, int sum, int cnt, int weight)

{

    if (weight > vMax) return ;

    if (cnt == k) ans = (sum > ans ? sum : ans);

    else

    {

        for (int i = cur+1; i <= n; ++i) if (vis[i] == false)

        {

            vis[i] = true;

            dfs(i, sum+v[i], cnt+1, weight+w[i]);

            vis[i] = false;

        }

    }

}



void init(void)

{

    scanf("%d%d", &n, &k);

    for (int i = 1; i <= n; ++i)

        scanf("%d%d", &v[i], &w[i]);

    scanf("%d", &vMax);

    memset(vis, false, sizeof(vis));

}



void solve(void)

{

    ans = 0;

    dfs(0, 0, 0, 0);

    printf("%d\n", ans);

}



int main()

{

    int T;



    scanf("%d", &T);

    while (T--)

    {

        init();

        solve();

    }



    return 0;

}

/**/