BZOJ 1227 [SDOI2009]虔诚的墓主人 - 扫描线

Solution

离散化 扫描线， 并用 $rest[i]$ 和 $cnt[i]$ 记录 第$i$列 总共有 $cnt[i]$棵常青树， 还有$rest[i]$ 没有被扫描到。

那么 第$i$ 列的方案数 为 $C(rest[i], k) * C(cnt[i]-rest[i], k)$。 乘上行上的方案数 并加入答案。

 

需要注意组合数要预处理， 我直接算发现$k > 2$就会WA。

Code

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define rd read()
#define R register
using namespace std;

const int N = 2e5 + 5;

int cnt[N], sum[N], n, r, l, k, rest[N];
int vis[11], ans;
int tot_x, tot_y, X[N], Y[N], c[N][15];

struct node {
    int x, y;
}pt[N];

vector<node> q[N];

inline int read() {
    R int X = 0, p = 1; char c = getchar();
    for (; c > '9' || c < '0'; c = getchar())
        if (c == '-') p = -1;
    for (; c >= '0' && c <= '9'; c = getchar())
        X = X * 10 + c - '0';
    return X * p;
}

inline int lowbit(int x) {
    return x & -x;
}

inline void add(R int x, int d) {
    for (;x <= tot_x; x += lowbit(x))
        sum[x] += d;
}

inline int query(R int x) {
    int re = 0;
    for (; x; x -= lowbit(x))
        re += sum[x];
    return re;
}

inline int fd_x(R int x) {
    return lower_bound(X + 1, X + 1 + tot_x, x) - X;
}

inline int fd_y(R int y) {
    return lower_bound(Y + 1, Y + 1 + tot_y, y) - Y;
}

inline int cmp(const node &A, const node &B) {
    return A.y == B.y ? A.x < B.x : A.y < B.y;
}

inline int C(int x) {
    return c[x][k];
}

int work(int x) {
    int len = q[x].size(), re = 0;
    for (R int j = k - 1; j <= len - k - 1; ++j) {
        int L = q[x][j].x, r = q[x][j + 1].x;
        int tmp = query(r - 1) - query(L);
        re += tmp * C(j + 1) * C(len - j - 1);
    }
    for (R int j = 0; j < len; ++j) {
        int tmp = C(rest[q[x][j].x]) * C(cnt[q[x][j].x] - rest[q[x][j].x]);
        add(q[x][j].x, -tmp);
        rest[q[x][j].x]--;
        tmp = C(rest[q[x][j].x]) * C(cnt[q[x][j].x] - rest[q[x][j].x]);
        add(q[x][j].x, tmp);
    }
    return re;
}

void init() {
    c[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
        c[i][0] = 1;
        for (int j = 1; j <= min(k, i); ++j)
            c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
    }
}

int main()
{
    r = rd, l = rd; n = rd;
    for (R int i = 1; i <= n; ++i) {
        pt[i].x = rd, pt[i].y = rd;
        X[++tot_x] = pt[i].x;
        Y[++tot_y] = pt[i].y;
    }
    k = rd;
    init();
    sort(X + 1, X + 1 + tot_x);
    sort(Y + 1, Y + 1 + tot_y);
    tot_x = unique(X + 1, X + 1 + tot_x) - X - 1;
    tot_y = unique(Y + 1, Y + 1 + tot_y) - Y - 1;
    sort(pt + 1, pt + 1 + n, cmp);
    for (R int i = 1; i <= n; ++i) {
        pt[i].x = fd_x(pt[i].x);
        pt[i].y = fd_y(pt[i].y);
        cnt[pt[i].x]++;
        q[pt[i].y].push_back(pt[i]);
    }
    for (R int i = 1; i <= tot_x; ++i)
        rest[i] = cnt[i];
    for (R int i = 1; i <= tot_y; ++i)
        ans += work(i);
    printf("%d\n", ans & 0x7fffffff);
}