HDU 4160 - Dolls (最大匹配)
题意
大的布娃娃可以装小的,只能装一个。求最后还剩下多少
思路
一对一,二分图最大匹配。减去匹配数就是剩下的。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 2000 + 10;
const int MOD = 1e9 + 7;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
struct EDGE
{
int from, to;
};
struct BIMATCHING
{
int vis[MAXN], link[MAXN], head[MAXN], next[MAXN * MAXN];
int nrem;
vector<EDGE> edges;
int DFS(int u)
{
for (int i = head[u]; i != -1; i = next[i])
{
int &v = edges[i].to;
if (!vis[v])
{
vis[v] = 1;
if (link[v] == -1 || DFS(link[v]))
{
link[v] = u;
return 1;
}
}
}
return 0;
}
int hungary()
{
int res = 0;
for (int i = 1; i <= nrem; i++)
{
MS(vis, 0);
res += DFS(i);
}
return res;
}
void add_edge(int from, int to)
{
edges.PB((EDGE){from, to});
int m = SZ(edges);
next[m - 1] = head[from];
head[from] = m - 1;
}
void init(int nrem)
{
this->nrem = nrem;
MS(head, -1); MS(link, -1);
edges.clear();
}
}hun;
struct DOLL
{
int a, b, c;
}dol[MAXN];
int main()
{
//ROP;
int n, i, j;
while (scanf("%d", &n), n)
{
hun.init(n);
int a, b, c;
for (i = 1; i <= n; i++)
scanf("%d%d%d", &dol[i].a, &dol[i].b, &dol[i].c);
for (i = 1; i <= n; i++)
for (j = 1; j <= n; j++)
if (dol[i].a > dol[j].a && dol[i].b > dol[j].b && dol[i].c > dol[j].c) hun.add_edge(i, j);
printf("%d\n", n - hun.hungary());
}
return 0;
}