計蒜客習題-俄羅斯套娃 最小路徑覆蓋

套娃和矩形嵌套一樣是典型的DAG

把嵌套關係用有向邊表示可以得到一個DAG

可以把DAG上的點i拆成兩個點 i 和 i' 分別放在兩個集合中 如果原圖i-j有邊 二分圖裏就連一條i-j'的邊 跑匈牙利求解

最小路徑覆蓋數==DAG的點數n（拆點前）-最大匹配數（拆點後的新圖）   即答案

 

//拆點保證不重複
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N = 507;
struct edge {
	int v, next;
}e[1000001];
int p[N<<1], eid = 0;
struct tw {
	int x, y, z;
}pool[N<<1];	//要拆點
int n;
inline int read() {
	char c = getchar();
	int s = 0, f = 1;
	while (c<'0' || c>'9') {
		if (c == '-') f = -1; c = getchar();
	}
	while (c >= '0'&&c <= '9') {
		s *= 10;
		s += c - 48;
		c = getchar();
	}
	return s*f;
}
inline void ins(int u,int v){
	e[eid].v = v; e[eid].next = p[u]; p[u] = eid++;
}
inline void init() { memset(p, -1, sizeof(p)); eid = 0; }
inline bool check(tw i, tw j) {
	return (i.x > j.x&&i.y > j.y&&i.z > j.z);
}
int link[N << 1];
bool vis[N << 1];
bool dfs(int u) {
	int v;
	for (int i = p[u]; ~i; i = e[i].next) {
		v = e[i].v;
		if (!vis[v]) {
			vis[v] = true;
			if (link[v] == -1 || dfs(link[v])){
				link[v] = u;
				return true;
			}
		}
	}
	return false;
}
int hungary() {
	int res = 0;
	memset(link, -1, sizeof(link));
	for (int i = 1; i <= n; i++)
	{
		memset(vis, 0, sizeof(vis));
		res+=dfs(i);
	}
	return res;
}
int main() {
	n = read();
	init();
	for (int i = 1; i <= n; i++)
	{
		pool[i].x = read(); pool[i].y = read(); pool[i].z = read();
	}
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++) {
				if (check(pool[i], pool[j])) ins(i, j+n);//i>j
				else if (check(pool[j], pool[i])) ins(j, i+n);
			}
		printf("%d", n-hungary());	//最小路徑覆蓋爲DAG點數-匹配數
		//getchar();
		return 0;
}