2015-2016 Northwestern European Regional Contest (NWERC 2015)

训练时间：2019-04-05

一场读错三个题，队友恨不得手刃了我这个坑B。

A I J 简单，不写了。

C - Cleaning Pipes (Gym - 101485C)

对于有公共点的管道建边，然后染色判是否是二分图。

注意线段判相交的时候，除了两个线段交于起点之外，都要视为相交。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn =  + ;
const double eps = 1e-;

int dcmp(double x)
{
    if (x > eps) return ;
    return x < -eps ? - : ;
}

struct Point
{
    double x, y;
    Point() {}
    Point(double _x, double _y) { x = _x, y = _y; }
    Point operator - (const Point &b) const
    {
        return Point(x-b.x, y-b.y);
    }

    bool operator == (const Point& b) const
    {
        return dcmp(x - b.x) ==  && dcmp(y - b.y) == ;
    }
};
typedef Point Vector;
struct Seg
{
    Point st, ed;
    Seg() {};
    Seg(Point a, Point b) { st = a, ed = b; }
};

double Cross(Vector a, Vector b) { return a.x*b.y - a.y*b.x; }
double Dot(Vector a, Vector b) { return a.x*b.x + a.y*b.y; }

bool OnSegment(Point p, Point a1, Point a2)
{
    return dcmp(Cross(a1-p, a2-p)) ==  && dcmp(Dot(a1-p, a2-p)) < ;
}

bool SegmentIntersection(Point a1, Point a2, Point b1, Point b2)
{
    if (a1 == b1 || a1 == b2 || a2 == b1 || a2 == b2) return true;

    if (OnSegment(a1, b1, b2) || OnSegment(a2, b1, b2)
        || OnSegment(b1, a1, a2) || OnSegment(b2, a1, a2)) return true;

    double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1),
            c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);
    return dcmp(c1) * dcmp(c2) <  && dcmp(c3) * dcmp(c4) < ;
}

vector<int> v[maxn];
int vis[maxn];

void build(int x, int y)
{
    v[x].push_back(y); v[y].push_back(x);
}

bool dfs(int x)
{
    for (int y : v[x])
    {
        if (vis[y] == vis[x]) return false;
        if (vis[y] == -)
        {
            vis[y] = !vis[x];
            if (!dfs(y)) return false;
        }
    }
    return true;
}

Point st[maxn];
Seg s[maxn];
int n, p;
int main()
{
    scanf("%d%d", &n, &p);
    for (int i = ; i <= n; i++) scanf("%lf%lf", &st[i].x, &st[i].y);
    for (int i = ; i <= p; i++)
    {
        double x, y;
        int id;
        scanf("%d%lf%lf", &id, &x, &y);
        s[i] = Seg(st[id], Point(x, y));
    }

    for (int i = ; i <= p; i++)
        for (int j = i+; j <= p; j++)
        {
            if (s[i].st == s[j].st) continue;
            if (SegmentIntersection(s[i].st, s[i].ed, s[j].st, s[j].ed)) build(i, j);
        }

    memset(vis, -, sizeof(vis));

    for (int i = ; i <= p; i++)
        if (vis[i] == -)
        {
            vis[i] = ;
            if (!dfs(i)) return printf("impossible\n"), ;
        }

    printf("possible\n");
}

D - Debugging (Gym - 101485D)

对于任何一个n行的代码块，可以由长度为n/2 n/3 n/4 ... n/n的代码块的答案更新。所以可以用dp做。

然而这个题没找到很好的递推顺序，窃以为递推为O(n^2)的。

如有合理递推方式，欢迎并感激指出！

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e6 + ;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-;

LL dp[maxn];

int n, r, p;

LL d(int n)
{
    if (n == ) return ;
    if (dp[n] != INF) return dp[n];
    for (int i = ; i < n; i++)
        dp[n] = min(dp[n], d( (int)ceil(1.0*n/(i+)) )+1ll*p*i+r);
    return dp[n];
}

int main()
{
    scanf("%d%d%d", &n, &r, &p);
    for (int i = ; i <= n; i++) dp[i] = INF;
    printf("%lld\n", d(n));
}

E - Elementary Math (Gym - 101485E)

把每组数字三种运算的结果计算出来，然后把结果和每组数字做一下二分图匹配。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = * + ;
const int maxm = * + ;

int n;
map<LL, int> M;
LL x[maxn], y[maxn];
LL tmp[maxn];
int v[maxm], last[maxm], nxt[maxm];
int lnk[maxn], vis[maxn], ans[maxn];
int cnt, tot;

void build(int x, int y)
{
    ++tot;
    v[tot] = y;
    nxt[tot] = last[x];
    last[x] = tot;
}

bool dfs(int x)
{
    for (int i = last[x]; i; i = nxt[i])
    {
        int y = v[i];
        if (!vis[y])
        {
            vis[y] = ;
            if (lnk[y] == - || dfs(lnk[y]))
            {
                lnk[y] = x;
                ans[x] = y;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    int res = ;
    memset(lnk, -, sizeof(lnk));
    for (int i = ; i <= n; i++)
    {
        memset(vis, , sizeof(vis));
        if (dfs(i)) res++;
    }
    return res;
}

char get(LL sum, int id)
{
    if (sum == x[id] + y[id]) return '+';
    if (sum == x[id] - y[id]) return '-';
    if (sum == x[id] * y[id]) return '*';
}

int main()
{
    scanf("%d", &n);
    cnt = n;
    for (int i = ; i <= n; i++)
    {
        scanf("%lld%lld", &x[i], &y[i]);
        LL a = x[i] + y[i], b = x[i] - y[i], c = x[i] * y[i];

        if (!M.count(a)) M[a] = ++cnt;
        build(i, M[a]); tmp[M[a]] = a;

        if (!M.count(b)) M[b] = ++cnt;
        build(i, M[b]); tmp[M[b]] = b;

        if (!M.count(c)) M[c] = ++cnt;
        build(i, M[c]); tmp[M[c]] = c;
    }

    if (hungary() != n) return printf("impossible\n"), ;

    for (int i = ; i <= n; i++)
        printf("%lld %c %lld = %lld\n", x[i], get(tmp[ans[i]], i), y[i], tmp[ans[i]]);
}

G - Guessing Camels (Gym - 101485G)

求一个三位偏序。

CDQ分治裸题，然而我不会。

看题解发现可以两两求二维偏序求出不合法的数目来，然后用总数目减去不合法数目。

二维偏序可以把一个序列的先后顺序映射到另一个中，然后求逆序对。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 2e5 + ;

int n;
int a[maxn], b[maxn], c[maxn];
int T[maxn], A[maxn];
LL cnt = ;

void merge_sort(int x, int y)
{
    if (y-x > )
    {
        int m = x + (y-x)/;
        int p = x, q = m, i = x;
        merge_sort(x, m);
        merge_sort(m, y);
        while(p < m || q < y)
        {
            if (q >= y || (p < m && A[p] <= A[q])) T[i++] = A[p++];
            else T[i++] = A[q++], cnt += m-p;
        }
        for (i = x; i < y; i++) A[i] = T[i];
    }
}

void mapping(int a[], int b[])
{
    int m[maxn];
    for (int i = ; i <= n; i++) m[a[i]] = i;
    for (int i = ; i <= n; i++) A[i] = m[b[i]];
    merge_sort(, n+);
}

int main()
{
    scanf("%d", &n);
    for (int i = ; i <= n; i++) scanf("%d", &a[i]);
    for (int i = ; i <= n; i++) scanf("%d", &b[i]);
    for (int i = ; i <= n; i++) scanf("%d", &c[i]);

    mapping(a, b);
    mapping(a, c);
    mapping(b, c);

    printf("%lld\n", (1ll * n * (n-) - cnt) / );
}