2020杭电多校第二场

1001.Total Eclipse

每次选择最大连通块，将所有值同时减小，减小到零断开。

换种思路，将所有数从大到小排序，依次将点加入图，加入后将所有值的权值降至下一个数的大小

此时答案为 (b[i]-b[i+1])*当前图的连通块个数，对每次加点维护一个连通块个数即可

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

struct point
{
    int id, w;
    bool operator<(point &f) const { return w > f.w; }
} a[100010];
vector<int> son[100010];
int n, m, u, v;
ll ans, block;
int pre[100010];
int find(int x) { return x == pre[x] ? x : pre[x] = find(pre[x]); }
void solve()
{
    cin >> n >> m;
    rep(i, 1, n)
    {
        cin >> a[i].w;
        a[i].id = i;
        son[i].clear();
        pre[i] = i;
    }
    rep(i, 1, m)
    {
        cin >> u >> v;
        if (a[u].w < a[v].w)
            son[u].push_back(v);
        else
            son[v].push_back(u);
    }
    sort(a + 1, a + n + 1);
    a[n + 1].w = 0;
    ans = block = 0;
    rep(i, 1, n)
    {
        int root = find(a[i].id);
        block++;
        for (vector<int>::iterator it = son[a[i].id].begin(); it != son[a[i].id].end(); it++)
        {
            int root2 = find(*it);
            if (root2 != root)
            {
                block--;
                pre[root2] = root;
            }
        }
        ans += block * (a[i].w - a[i + 1].w);
    }
    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();
    }
}

1006.The Oculus

修改的那一位斐波拉契数为f[k]，满足A*B==C+f[k]

在f[1],f[2]...f[2000000]%mod不同余的时候就能找到唯一的k

题解mod取2⁶⁴，不太明白为什么

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

unsigned long long f[2000010], a[3];
int n, x;
void solve()
{
    a[0] = a[1] = a[2] = 0;
    rep(i, 0, 2)
    {
        cin >> n;
        rep(j, 1, n)
        {
            cin >> x;
            if (x)
                a[i] += f[j];
        }
    }
    for (int i = 1;; i++)
        if (a[0] * a[1] - a[2] == f[i])
        {
            cout << i << endl;
            break;
        }
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    f[1] = 1;
    f[2] = 2;
    rep(i, 3, 2000000) f[i] = f[i - 2] + f[i - 1];

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

1010.Lead of Wisdom

暴搜。。。我去除了比某一同类装备4个属性都低的装备

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

typedef struct clo
{
    int a, b, c, d;
} cloth;
int n, k, A, B, C, D, T;
ll ans;
vector<cloth> p[55];
void dfs(int x, int a, int b, int c, int d)
{
    while (x <= k && p[x].empty())
        x++;
    if (x > k)
    {
        ans = max(ans, 1ll * a * b * c * d);
        return;
    }
    for (vector<cloth>::iterator it = p[x].begin(); it != p[x].end(); it++)
        dfs(x + 1, a + (*it).a, b + (*it).b, c + (*it).c, d + (*it).d);
}
inline void solve()
{

    cin >> n >> k;
    rep(i, 1, k) p[i].clear();
    ans = 100000000;
    rep(i, 1, n)
    {
        cin >> T >> A >> B >> C >> D;
        int flag = 1;
        for (vector<cloth>::iterator it = p[T].begin(); it != p[T].end(); it++)
        {
            if (A <= (*it).a && B <= (*it).b && C <= (*it).c && D <= (*it).d)
            {
                flag = 0;
                break;
            }
        }
        if (flag)
            p[T].push_back({A, B, C, D});
    }
    dfs(1, 100, 100, 100, 100);
    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();
    }
}