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 + ;

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

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

 vector<node> q[N];

 inline int read() {
     R int X = , p = ; char c = getchar();
     for (; c > '' || c < ''; c = getchar())
         if (c == '-') p = -;
     for (; c >= '' && c <= ''; c = getchar())
         X = X *  + c - '';
     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 = ;
     for (; x; x -= lowbit(x))
         re += sum[x];
     return re;
 }

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

 inline int fd_y(R int y) {
     return lower_bound(Y + , Y +  + 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 = ;
     for (R int j = k - ; j <= len - k - ; ++j) {
         int L = q[x][j].x, r = q[x][j + ].x;
         int tmp = query(r - ) - query(L);
         re += tmp * C(j + ) * C(len - j - );
     }
     for (R int j = ; 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[][] = ;
     for (int i = ; i <= n; ++i) {
         c[i][] = ;
         for (int j = ; j <= min(k, i); ++j)
             c[i][j] = c[i - ][j - ] + c[i - ][j];
     }
 }

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