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
" , 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());
}