JZOJ 5354. 【NOIP2017提高A组模拟9.9】导弹拦截

题目

如题

分析

第一问很简单， \(dp\) 即可（得先排序）

第二问很经典，最小路径覆盖问题，最大流解决 \(n-Maxflow\)

\(Code\)

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;

const int N = 1005;
int n , f[N] , tot = 1 , dep[2 * N] , cur[2 * N] , h[2 * N] , S = 0 , T = 2 * N - 9;
struct node{
	int x , y , z;
}a[N];
struct edge{
	int to , nxt , w;
}e[2 * N * N];

bool cmp(node x , node y)
{
	return (x.x < y.x ? 1 : (x.x == y.x ? (x.y < y.y ? 1 : (x.y == y.y ? x.z < y.z : 0)) : 0));
}
inline void add(int x , int y , int z){e[++tot] = edge{y , h[x] , z} , h[x] = tot;}

queue<int> q;
int bfs()
{
	while (!q.empty()) q.pop();
	for(register int i = S; i <= T; i++) cur[i] = h[i] , dep[i] = 0;
	dep[S] = 1 , q.push(S);
	while (!q.empty())
	{
		int now = q.front(); q.pop();
		for(register int i = h[now]; i; i = e[i].nxt)
		{
			int v = e[i].to;
			if (dep[v] || e[i].w == 0) continue;
			dep[v] = dep[now] + 1 , q.push(v);
		}
	}
	return dep[T];
}

int dfs(int x , int fa , int mi)
{
	if (x == T || mi <= 0) return mi;
	int flow = 0;
	for(register int i = cur[x]; i; i = e[i].nxt)
	{
		cur[x] = i;
		int v = e[i].to;
		if (e[i].w == 0 || v == fa || dep[x] + 1 != dep[v]) continue;
		int f = dfs(v , x , min(mi , e[i].w));
		if (f <= 0) continue;
		flow += f , mi -= f , e[i].w -= f , e[i ^ 1].w += f;
		if (mi <= 0) break;
	}
	return flow;
}

int dinic()
{
	int flow = 0;
	while (bfs()) flow += dfs(S , -1 , n);
	return flow;
}

int main()
{
	freopen("missile.in" , "r" , stdin);
	freopen("missile.out" , "w" , stdout);
	scanf("%d" , &n);
	for(register int i = 1; i <= n; i++) scanf("%d%d%d" , &a[i].x , &a[i].y , &a[i].z);
	sort(a + 1 , a + n + 1 , cmp);
	for(register int i = 1; i <= n; i++)
	{
		f[i] = 1;
		for(register int j = 1; j < i; j++)
		if (a[j].x < a[i].x && a[j].y < a[i].y && a[j].z < a[i].z && f[j] + 1 > f[i]) f[i] = f[j] + 1;
	}
	int ans = 0;
	for(register int i = 1; i <= n; i++) ans = max(ans , f[i]);
	printf("%d\n" , ans);
	for(register int i = 1; i <= n; i++)
		add(S , i , 1) , add(i , S , 0) , add(i + n , T , 1) , add(T , i + n , 0);
	for(register int i = 1; i <= n; i++)
		for(register int j = 1; j <= n; j++)
		if (i != j && a[j].x > a[i].x && a[j].y > a[i].y && a[j].z > a[i].z)
			add(i , j + n , 1) , add(j + n , i , 0);
	printf("%d" , n - dinic());
}