2020杭电多校第三场

1004.Tokitsukaze and Multiple

求和为p的倍数的块的最大数量，我用了比较暴力的做法，好像有dp的做法？

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)

int n, p;
int f, ans, flag;
inline void solve()
{
    cin >> n >> p;
    vector<int> a;
    ans = 0;
    rep(i, 1, n)
    {
        cin >> f;
        flag = 0;
        if (f % p == 0)
            flag = 1;
        else
            for (vector<int>::iterator it = a.begin(); it != a.end(); it++)
            {
                *it += f;
                if (*it % p == 0)
                {
                    flag = 1;
                    break;
                }
            }
        if (flag)
        {
            ans++;
            a.clear();
        }
        else
            a.push_back(f);
    }
    cout << ans << endl;
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

1005.Little W and Contest

并查集维护块的数量，每合并一个块减去对应产生的贡献。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)

int mod = 1e9 + 7;

int pre[100010];
int find(int x) { return x == pre[x] ? x : pre[x] = find(pre[x]); }
int n, a[100010];
int u, v;
int cnt1, cnt2, s1[100010], s2[100010];
ll ans;
inline void solve()
{
    cin >> n;
    cnt1 = cnt2 = 0;
    rep(i, 1, n)
    {
        cin >> a[i];
        pre[i] = i;
        if (a[i] == 1)
        {
            s1[i] = 1;
            s2[i] = 0;
            cnt1++;
        }
        else
        {
            s1[i] = 0;
            s2[i] = 1;
            cnt2++;
        }
    }
    ans = 1ll * cnt2 * (cnt2 - 1) * (cnt2 - 2) / 6 + 1ll * cnt2 * (cnt2 - 1) * cnt1 / 2;
    cout << ans % mod << endl;
    rep(i, 1, n - 1)
    {
        cin >> u >> v;
        u = find(u);
        v = find(v);
        ans -= 1ll * s2[u] * s2[v] * (cnt2 - s2[u] - s2[v]);
        ans -= 1ll * s2[u] * s2[v] * (cnt1 - s1[u] - s1[v]);
        ans -= 1ll * s1[u] * s2[v] * (cnt2 - s2[u] - s2[v]);
        ans -= 1ll * s2[u] * s1[v] * (cnt2 - s2[u] - s2[v]);
        pre[u] = v;
        s1[v] += s1[u];
        s2[v] += s2[u];
        cout << ans % mod << endl;
    }
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

1007.Tokitsukaze and Rescue

边权随机的情况下，最短路的边数很少？

求出最短路，再依次删除这条路所有边，递归求解。

复杂度为求最短路的复杂度*最短路边数的k次方

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)

int n, k;
int mp[55][55], u, v, w;
int d[55], vis[55], f[10][55];
int ans;

void dfs(int step)
{
    rep(i, 0, n)
    {
        d[i] = 100000000;
        vis[i] = 0;
        f[step][i] = 0;
    }
    d[1] = 0;
    rep(i, 1, n)
    {
        int k = 0;
        rep(j, 1, n) if (!vis[j] && d[j] < d[k]) k = j;
        vis[k] = 1;
        rep(j, 1, n) if (!vis[j] && d[k] + mp[j][k] < d[j])
        {
            d[j] = d[k] + mp[j][k];
            f[step][j] = k;
        }
    }
    if (step == 0)
    {
        ans = max(d[n], ans);
        return;
    }
    for (int i = n, j, k; i != 1; i = j)
    {
        j = f[step][i];
        k = mp[i][j];
        mp[i][j] = mp[j][i] = 100000000;
        dfs(step - 1);
        mp[i][j] = mp[j][i] = k;
    }
}
inline void solve()
{
    cin >> n >> k;
    rep(i, 1, n * (n - 1) / 2)
    {
        cin >> u >> v >> w;
        mp[u][v] = mp[v][u] = w;
    }
    ans = 0;
    dfs(k);
    cout << ans << endl;
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

1009.Parentheses Matching

将'*'换成 '('，')'或' '，求最短的，字典序最小的平衡字符串。

先将左右括号的个数整相等，再判断字符串合不合法，不合法再成对的改变*号

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)

char s[100010];
int cnt1, cnt2, num, sum[100010];
int tmp[100010], l, r, id;
inline void solve()
{
    cin >> s;
    int n = strlen(s);
    l = 1;
    cnt1 = cnt2 = r = 0;
    rep(i, 0, n - 1) if (s[i] == '(')
        cnt1++;
    else if (s[i] == ')')
        cnt2++;
    else tmp[++r] = i;
    if (abs(cnt2 - cnt1) > r)
    {
        puts("No solution!");
        return;
    }
    if (cnt1 > cnt2)
        rep(i, 1, cnt1 - cnt2) s[tmp[r--]] = ')';
    if (cnt1 < cnt2)
        rep(i, 1, cnt2 - cnt1) s[tmp[l++]] = '(';
    num = 0;
    rep(i, 0, n - 1)
    {
        if (s[i] == '*')
            continue;
        else if (s[i] == '(')
            num++;
        else if (s[i] == ')')
            num--;
        if (num < 0)
        {
            if (l >= r || tmp[l] > i)
            {
                puts("No solution!");
                return;
            }
            s[tmp[l++]] = '(';
            s[tmp[r--]] = ')';
            num = 0;
        }
    }
    if (num != 0)
    {
        puts("No solution!");
        return;
    }
    rep(i, 0, n - 1) if (s[i] == '*') continue;
    else putchar(s[i]);
    puts("");
}
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }
}