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。

# 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;
}

/**/