HDU - 4160 最小路径覆盖

题意：

思路：

代码：

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 505;
const int MAXM = 505;

struct node {
int x, y, z;
bool operator < (const node &b) const {
if (x == b.x) {
if (y == b.y) return z < b.z;
return y < b.y;
}
return x < b.x;
}
}a[MAXN];

struct BgMaxMatch {

int nl, nr;
int l[MAXN], r[MAXM];
bool vis[MAXN];
vector <int> G[MAXN];  //这里注意要看是按左右哪边建图，也有可能是MAXM

void init(int nl, int nr) {
this -> nl = nl;
this -> nr = nr;
for (int i = 1; i <= nl; i++) G[i].clear();
memset(vis, false, sizeof(vis));
}

void AddEdge(int u, int v) {
G[u].push_back(v);
}

bool dfs(int u){
int cnt = G[u].size();
for (int i = 0; i < cnt; i++) {
int v = G[u][i];
if (!vis[v]) {
vis[v] = true;
if(l[v] == -1 || dfs(l[v]))	{
l[v] = u; r[u] = i;
return true;
}
}
}
return false;
}

int MaxMatch() {
memset(l, -1, sizeof(l));
memset(r, -1, sizeof(r));
int ans = 0;
for (int i = nl; i >= 1; i--) {
memset(vis, false, sizeof(vis));
ans += dfs(i);
}
return ans;
}
} solver;

int main() {
int n;
while (scanf("%d", &n), n) {
for (int i = 1; i <= n; i++)
scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].z);
sort (a + 1, a + 1 + n);
solver.init(n, n);
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
if (a[i].x < a[j].x && a[i].y < a[j].y && a[i].z < a[j].z)
solver.AddEdge(i, j);
}
}
printf("%d\n", n - solver.MaxMatch());
}
return 0;
}