CTU Open Contest 2017

B - Pond Cascade

优先队列维护这个水池需要多少时间 或者 直接扫一遍。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 100000 + 10;
const int INF = 2e9;

LL a[maxn], sum[maxn];

int main()
{
    int n;
    LL flow;
    while(~scanf("%d%lld", &n, &flow))
    {
        for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);

        double t1 = INF, sum1 = 0, t2 = -INF, sum2 = 0;
        for (int i = n; i >= 1; i--)
            sum1 += a[i], t1 = min(t1, sum1*1.0/(n-i+1)/flow);
        for (int i = 1; i <= n; i++)
            sum2 += a[i], t2 = max(t2, sum2*1.0/i/flow);

        printf("%.8lf %.8lf\n", t1, t2);
    }
}

D - Equinox Roller Coaster

N个点，找出一个可以构成的最大的正方形，并且正方形的边都与坐标轴平行。

对于每个X/Y坐标建一个表。对于每个点，枚举它X/Y坐标对应的表中所有的点（选size较小的那一维），然后判断剩下的另外两个点是否存在。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
using namespace std;
const int maxn = 100000 + 100;
map<int, set<int> > px, py;
int x[maxn], y[maxn];

int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        px.clear(), py.clear();
        for (int i = 1; i <= n; i++)
        {
            scanf("%d%d", &x[i], &y[i]);
            px[x[i]].insert(y[i]), py[y[i]].insert(x[i]);
        }

        int ans = 0;
        for (int i = 1; i <= n; i++)
        {
            if (px[x[i]].size() < py[y[i]].size())
                for (auto j = px[x[i]].begin(); j != px[x[i]].end(); j++)
                {
                    if ((*j) == y[i]) continue;
                    int dis = (*j) - y[i];
                    if (px[x[i] + dis].count(y[i]) && px[x[i] + dis].count(*j))
                        ans = max(ans, dis);
                }
            else
                for (auto j = py[y[i]].begin(); j != py[y[i]].end(); j++)
                {
                    if ((*j) == x[i]) continue;
                    int dis = (*j) - x[i];
                    if (py[y[i] + dis].count(x[i]) && py[y[i] + dis].count(*j))
                        ans = max(ans, dis);
                }
        }
        printf("%d\n", ans);
    }
}