PKU 3041 - Asteroids (最小点覆盖 & 匈牙利)
题意
有一支激光枪,可以干掉一行或者一列,问最少需要多少枪。
思路
因为一枪可以干掉一行,所以选择x、y建图,求最小点覆盖,就是求最大匹配数。
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 = 1000 + 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;
int main()
{
//ROP;
int n, k, i, j;
while (~scanf("%d%d", &n, &k))
{
hun.init(n);
while (k--)
{
int x, y;
scanf("%d%d", &x, &y);
hun.add_edge(x, y);
}
printf("%d\n", hun.hungary());
}
return 0;
}