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  • 「luogu3810」陌上花开

    「luogu3810」陌上花开

    传送门
    三维偏序, ( ext{CDQ}) 分治板子题。
    判重的地方注意一下,别的就都是板子了。
    参考代码:

    #include <algorithm>
    #include <cstdio>
    #define rg register
    #define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
    using namespace std;
    template < class T > inline void read(T& s) {
    	s = 0; int f = 0; char c = getchar();
    	while ('0' > c || c > '9') f |= c == '-', c = getchar();
    	while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
    	s = f ? -s : s;
    }
    
    const int _ = 1e5 + 5, __ = 2e5 + 5;
    
    int n, k, tr[__], ans[_];
    
    inline int lb(int x) { return x & -x; }
    
    inline void update(int x, int v)
    { for (rg int i = x; i <= k; i += lb(i)) tr[i] += v; }
    
    inline int query(int x)
    { int res = 0; for (rg int i = x; i >= 1; i -= lb(i)) res += tr[i]; return res; }
    
    struct node { int a, b, c, f, w; } t[_], tt[_];
    inline bool cmp(const node& x, const node& y)
    { return x.a != y.a ? x.a < y.a : (x.b != y.b ? x.b < y.b : x.c < y.c); }
    
    inline void CDQ(int l, int r) {
    	if (l == r) return ;
    	int mid = (l + r) >> 1;
    	CDQ(l, mid), CDQ(mid + 1, r);
    	int i = l, j = mid + 1, p = l;
    	while (i <= mid && j <= r) {
    		if (t[i].b <= t[j].b)
    			update(t[i].c, t[i].w), tt[p++] = t[i++];
    		else
    			t[j].f += query(t[j].c), tt[p++] = t[j++];
    	}
    	while (i <= mid) update(t[i].c, t[i].w), tt[p++] = t[i++];
    	while (j <= r) t[j].f += query(t[j].c), tt[p++] = t[j++];
    	for (rg int i = l; i <= mid; ++i) update(t[i].c, -t[i].w);
    	for (rg int i = l; i <= r; ++i) t[i] = tt[i];
    }
    
    int main() {
    	read(n), read(k);
    	for (rg int i = 1; i <= n; ++i)
    		read(t[i].a), read(t[i].b), read(t[i].c), t[i].w = 1;
    	sort(t + 1, t + n + 1, cmp);
    	int cnt = 1;
    	for (rg int i = 2; i <= n; ++i) {
    		if (t[i].a == t[cnt].a && t[i].b == t[cnt].b && t[i].c == t[cnt].c) ++t[cnt].w;
    		else t[++cnt] = t[i];
    	}
    	CDQ(1, cnt);
    	for (rg int i = 1; i <= cnt; ++i) ans[t[i].f + t[i].w - 1] += t[i].w;
    	for (rg int d = 0; d < n; ++d) printf("%d
    ", ans[d]);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zsbzsb/p/12231675.html
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