Grakn Forces 2020

题目传送门

A. Circle Coloring

#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, a[110][3];
int flag;
int ans[110];
void dfs(int step) {
    if(step == n + 1) {
        if(a[n][ans[n]] == a[1][ans[1]]) return;
        rep(i, 1, n) cout << a[i][ans[i]] << " ";
        cout << endl;
        flag = 1;
    }
    if(flag) return;
    rep(k, 0, 2) {
        if(a[step][k] == a[step - 1][ans[step - 1]]) continue;
        ans[step] = k;
        dfs(step + 1);
        if(flag) break;
    }
    
}
 
inline void solve(int T)
{
    cin >> n;
    rep(k, 0, 2) rep(i, 1, n) cin >> a[i][k];
    flag = 0;
    dfs(1);
}
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
}

 

B. Arrays Sum

每一个B序列最多消除$k - 1$个上升项

特判一下$k$等于$0$的情况

#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 a[110], cnt;
inline void solve(int T)
{
    cin >> n >> k;
    k--;
    a[0] = cnt = -1;
    rep(i, 1, n){
        cin >> a[i];
        if(a[i] > a[i - 1]) cnt++;
    }
    if(cnt == 0) cout << "1" << endl;
    else if(k == 0) cout << "-1" << endl;
    else cout << ceil(1.0 * cnt / k) << endl;
}
 
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
}

 

C. Discrete Acceleration

模拟

#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, l;
int a[100010];
double s1[100010], s2[100010];
inline void solve(int T)
{
    cin >> n >> l;
    rep(i, 1, n) {
        cin >> a[i];
        s1[i] = s1[i - 1] + 1.0 * (a[i] - a[i - 1]) / i;
    }
    a[n + 1] = l;
    s2[n + 1] = 0;
    for(int i = n; i; i--) s2[i] = s2[i + 1] + 1.0 * (a[i + 1] - a[i]) / (n - i + 1);
    int tmp = n;
    rep(i, 1, n) if(s1[i] > s2[i]) {
        tmp = i - 1;
        break;
    }
    double len = a[tmp + 1] - a[tmp];
    if(s1[tmp] <= s2[tmp + 1]) len -= (tmp + 1) * (s2[tmp + 1] - s1[tmp]);
    else len -= (n - tmp + 1) * (s1[tmp] - s2[tmp + 1]); 
 
    printf("%.15lf\n", max(s1[tmp], s2[tmp + 1]) + len / (n + 2));
 
}
 
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    cin >> T;
    rep(i, 1, T) solve(i);
}

 

D. Searchlights

从大到小枚举一个方向需要改变的长度，再记录一下另一个方向需要改变的长度的最大值

#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, m;
int a[2010], b[2010], c[2010], d[2010];
vector<pair<int, int> > cnt;
bool cmp(pair<int, int> x, pair<int, int> y) {return x.first > y.first;}
inline void solve(int T)
{
    cin >> n >> m;
    rep(i, 1, n) cin >> a[i] >> b[i];
    rep(i, 1, m) cin >> c[i] >> d[i];
    rep(i, 1, n) rep(j, 1, m) if(a[i] <= c[j] && b[i] <= d[j]) cnt.push_back(make_pair(c[j] - a[i] + 1, d[j] - b[i] + 1));
    if(cnt.empty()) {
        cout << "0" << endl;
        return;
    }
    sort(cnt.begin(), cnt.end(), cmp);
    int num = cnt.size(), ans = cnt[0].first, mx = cnt[0].second;
    rep(i, 1, num - 1){
        if(cnt[i].first != cnt[i - 1].first) ans = min(cnt[i].first + mx, ans);
        mx = max(mx, cnt[i].second);
    }
    ans = min(ans, mx);
    cout << ans << endl;
}
 
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
 
    // freopen("in.txt", "r", stdin);
    // freopen("ans.txt", "w", stdans);
 
    int T = 1;
    // cin >> T;
    rep(i, 1, T) solve(i);
}