CF878C Tournament set 图论

题面

题面

题解

如果2个人可以互相战胜，那么我们连一条无向边，于是最后会剩下t个联通块，其中每对联通块之间都有严格的大小关系(a.max < b.min),因此我们每插入一个点就相当于合并一段连续的块，其中边界(无法继续合并的地方)是第一个严格比它小的和第一个严格比它大的。

#include<bits/stdc++.h>
using namespace std;
#define R register int
#define LL long long
#define ld double
#define AC 15
#define ac 101000

int n, K;
inline int read()
{
	int x = 0;char c = getchar();
	while(c > '9' || c < '0') c = getchar();
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x;
}

inline void upmin(int &a, int b) {if(b < a) a = b;}
inline void upmax(int &a, int b) {if(b > a) a = b;}

struct node{
	int maxn[AC], minn[AC], Size;
	
	friend bool operator < (node a, node b)
	{
		for(R i = 1; i <= K; i ++)
			if(a.maxn[i] > b.minn[i]) return 0;
		return 1;//只有b严格大于a才返回true，即Min(b) > Max(a)
	}
	
	inline void merge(node a)
	{
		Size += a.Size;
		for(R i = 1; i <= K; i ++)
			upmax(maxn[i], a.maxn[i]), upmin(minn[i], a.minn[i]);
	}
};
set <node> S;
set <node> :: iterator q[ac];

void pre()
{
	n = read(), K = read();
}

void work()
{
	for(R i = 1; i <= n; i ++)
	{
		node x;
		x.Size = 1;
		for(R j = 1; j <= K; j ++) x.maxn[j] = x.minn[j] = read();
		set <node> :: iterator last = S.lower_bound(x);
		set <node> :: iterator Next = S.upper_bound(x);
		set <node> :: iterator it;
		int tot = 0;
		for(it = last; it != Next; it ++) q[++ tot] = it;
		for(R j = 1; j <= tot; j ++) x.merge(*q[j]), S.erase(q[j]);//将中间的点不断合并进来并删除
		S.insert(x);
		printf("%d
", S.rbegin() -> Size);
	}
}

int main()
{
	freopen("in.in", "r", stdin);
	pre();
	work();
	fclose(stdin);
	return 0;
}