google kcikstart 2018 round c

Posted 2020-11-17 hxidkd

https://code.google.com/codejam/contest/4384486/dashboard#s=p0

A

题意

给定一个无向图，其中存在着唯一的一个环，求每个点到这个环的最短距离。

数据范围

≤ T ≤ 100.
1 ≤ xi ≤ N, for all i.
1 ≤ yi ≤ N, for all i.
xi ≠ yi, for all i.
(xi, yi) ≠ (xj, yj), for all i ≠ j.
The graph in which planets are nodes and tubes are edges is connected and has exactly one cycle.

small

3 ≤ N ≤ 30.

large

3 ≤ N ≤ 1000.

题解

显然关键在于找出这个环上的点。找出之后即使是大数据集也可以暴力BFS遍历每个点到这个换的最短距离(O(n^2))，或者可以反过来从环上bfs到所有点(O(n))。

比较暴力的找环方法是直接DFS遍历，然后list存储走过的点，当点的所有child dfs完之后将该点删掉。这样遇到重复的点时这之间的所有点就是该环。

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
vector<int> g[1005];
vector<int> path;
vector<int> circle;
bool visited[1005];
void dfs(int cur, int pa) {
    if (visited[cur] == true) {
        auto it = find(path.begin(), path.end(), cur);
        if (it != path.end()) {
            while (it != path.end()) {
                circle.push_back(*it);
                it++;
            }
        }
        return;
    }
    visited[cur] = true;
    path.push_back(cur);
    for (int &child : g[cur]) {
        if (child != pa) {
            dfs(child, cur);
        }
    }
    path.pop_back();
}
void solve() {
    int n;
    cin >> n;
    int x, y;
    for (int i = 1; i <= n; i++) {
        g[i].clear();
    }
    path.clear();
    circle.clear();
    memset(visited, false, sizeof(visited));
    for (int i = 1; i <= n; i++) {
        cin >> x >> y;
        g[x].push_back(y);
        g[y].push_back(x);
    }
    dfs(1, -1);
    vector<int> dis(n+1, n+1);
    queue<int> q;
    int temp = circle.size();
    for (int i = 0; i < temp; i++) {
        //cout << circle[i] << "
";
        dis[circle[i]] = 0;
        q.push(circle[i]);
    }
    int cur = 1;
    while (!q.empty()) {
        int sz = q.size();
        for (int i = 0; i < sz; i++) {
            int top = q.front();
            q.pop();
            for (int &child : g[top]) {
                if (dis[child] > cur) {
                    dis[child] = cur;
                    q.push(child);
                }
            }
        }
        cur++;
    }
    printf("%d",dis[1]);
    for (int i = 2; i <= n; i++) {
        printf(" %d", dis[i]);
    }
    printf("
");
}
int main() {
    freopen("A-large.in", "r", stdin);
    freopen("A-large.out", "w", stdout);
    int t;
    cin >> t;
    for (int tt = 1; tt <= t; tt++) {
        printf("Case #%d: ", tt);
        solve();
    }
    return 0;
}

B

题意

求从一个无向图中拆出的边能组成的凸边形个数。其中拆出的便不能有相同的端点。

数据范围

0 ≤ Li, j ≤ 1000 for all i, j.
Li, i = 0, for all i.
Li, j = Lj, i, for all i, j.

small

N = 6

large

6<=N<=15

题解

对小数据集，显然最多只能拆出三条边组成三角形，暴力即可。

大数据集不会，待补。

C

题意

给一个数组，对数组中的所有连续子数组按以下公式求k次和，然后求和最后所有的结果(modulo 1000000007)。
对连续子数组Aj, Aj+1, ..., Ak，第i次求和公式为 Aj × 1i + Aj+1 × 2i + Aj+2 × 3i + ... + Ak × (k-j+1)i.

样例：

In Sample Case #1, the Parameter Array is [3, 2]. All the contiguous subarrays are [3], [2], [3, 2].

For i = 1:

1-st Exponential-power of [3] = 3 × 11 = 3
1-st Exponential-power of [2] = 2 × 11 = 2
1-st Exponential-power of [3, 2] = 3 + 2 × 21 = 7
So POWER1 is 12.
For i = 2:

2-nd Exponential-power of [3] = 3 × 12 = 3
2-nd Exponential-power of [2] = 2 × 12 = 2
2-nd Exponential-power of [3, 2] = 3 + 2 × 22 = 11
So POWER2 is 16.
For i = 3:

3-rd Exponential-power of [3] = 3 × 13 = 3
3-rd Exponential-power of [2] = 2 × 13 = 2
3-rd Exponential-power of [3, 2] = 3 + 2 × 23 = 19
So POWER3 is 24.

So answer is 24 + 12 + 16 = 53.

数据范围

1 ≤ T ≤ 100.
1 ≤ x1 ≤ 1e5.
1 ≤ y1 ≤ 1e5
1 ≤ C ≤ 1e5.
1 ≤ D ≤ 1e5.
1 ≤ E1 ≤ 1e5.
1 ≤ E2 ≤ 1e5.
1 ≤ F ≤ 1e5.

small

1 ≤ N ≤ 100.
1 ≤ K ≤ 20.

large

1 ≤ N ≤ 1e6.
1 ≤ K ≤ 1e4

题解

首先根据xycdef生成数组A。
对于小数据集暴力即可。O(k*n^3)
对于大数据集，可以考虑每个数对最后结果的贡献，大力推推就可以出来了。

考虑第i次计算，则数组A中第j位计算的值为：A[j] * (N+1-j) * {1^i + 2^i + .... + j^i} ( j从1开始) 。直接枚举的话复杂度O（k * n^2lgk) ，当然第j^i求和可以累加，能降到O（knlgk) 。
另一种思路是调换一下求和顺序，则ans=sum_{i=1..j} {A[j] (N+1-j) * sum_{i=1..k}{1^i + 2^i + .... + j^i}}。其中m^i求和可以用等比数列公式，再通过累加可以降到O（k*lgk)。(lgk为求幂时的复杂度)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const ll mod = 1e9 + 7;
ll pw(ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b&1) ans = ans * a % mod;
        b /= 2;
        a = a * a % mod;
        //cout << ans << " " << a << "
";
    }
    return ans;
}
ll calc(ll a, ll b) {
    //cout << pw(a, b) << "
";
    ll ans = a * ((pw(a, b) - 1 + mod) % mod) % mod * pw(a-1, mod-2) % mod;
    //cout << ans <<"
";
}
ll solve() {
    ll n, k;
    cin >> n >>k;
    vector<ll> nums(n+1);
    ll x, y;
    cin >> x >> y;
    ll c,d,e1,e2,f;
    cin >> c >> d >> e1 >> e2 >> f;
    nums[1] = (x+y) % f;
    for (int i = 2; i <= n; i++) {
        ll tx = x, ty = y;
        x = (c*tx+d*ty+e1) %f;
        y = (d*tx+c*ty+e2) %f;
        nums[i] = (x+y) % f;
    }
    ll ans = nums[1]*n%mod*k%mod;
    ll sum = k;
    //cout << ans <<"
";
    for (int i = 2; i <= n; i++) {
        sum = (sum + calc(i, k)) % mod;
        //cout << sum << "
";
        ans = (ans + nums[i] * (n+1-i) % mod * sum % mod) % mod;
        //cout << i << " " << ans << "
";
    }
    return ans;
}
int main() {
    freopen("C-large-practice.in", "r", stdin);
    freopen("C-large.out", "w", stdout);
    int t;
    cin >> t;
    for (int tt = 1; tt <= t; tt++) {
        std::cerr << tt << "
";
        printf("Case #%d: %lld
", tt, solve());
    }
    return 0;
}