poj 2446 poj 1469 poj 1274 二分图最大匹配

poj 2446：

题意：

给一张m * n的图，其中挖掉一些点，然后用1*2的小纸片来覆盖这张图，问能否将这个图覆盖。

如下图：



解析：

将每个不为挖掉的点视为二分图的X点集，这个点的四个方向上的点视为Y点集，此时最大匹配数即为点的个数（想一想为什么）。

因为有重复匹配 1和2

2和1

这题用一个id数组来处理每个点的id，然后用匈牙利算法求最大匹配就ok了。

代码：

（邻接矩阵版）1552K 172MS

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <climits>
#include <cassert>
#define LL long long
#define lson lo, mi, rt << 1
#define rson mi + 1, hi, rt << 1 | 1

using namespace std;
const int maxn = 33 * 33;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double ee = exp(1.0);

int dir[][2] = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
bool maze[maxn][maxn];
vector<int> g[maxn];
bool vis[maxn];
int fr[maxn];
int m, n, k;
int cnt;
int id[maxn][maxn];

bool match(int v)
{
for (int i = 0; i < g[v].size(); i++)
{
int u = g[v][i];
if (!vis[u])
{
vis[u] = true;
if (fr[u] == -1 || match(fr[u]))
{
fr[u] = v;
return true;
}
}
}
return false;
}

int hungary()
{
int ret = 0;
memset(fr, -1, sizeof(fr));
for (int i = 1; i <= cnt; i++)
{
memset(vis, false, sizeof(vis));
if (match(i))
ret++;
}
return ret;
}

int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif // LOCAL
while (~scanf("%d%d%d", &m, &n, &k))
{
memset(maze, true, sizeof(maze));
for (int i = 0; i < k; i++)
{
int x, y;
scanf("%d%d", &y, &x);
maze[x][y] = false;
}
if ((m * n - k) % 2)
{
printf("NO\n");
}
else
{
cnt = 0;
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
if (maze[i][j])
{
id[i][j] = ++cnt;
}
}
}
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
if (maze[i][j])
{
for (int d = 0; d < 4; d++)
{
int ni = i + dir[d][0];
int nj = j + dir[d][1];
if (1 <= ni && ni <= m && 1 <= nj && nj <= n && maze[ni][nj])
{
g[id[i][j]].push_back(id[ni][nj]);
}
}
}
}
}
//cout << hungary() <<endl;
if (hungary() + k == m * n)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
}
}
return 0;
}

（邻接表） 1492K 94MS

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <climits>
#include <cassert>
#define LL long long
#define lson lo, mi, rt << 1
#define rson mi + 1, hi, rt << 1 | 1

using namespace std;
const int maxn = 33 * 33;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double ee = exp(1.0);

int dir[][2] = {{0, -1}, {0, 1}, {1, 0}, {-1, 0}};
bool maze[maxn][maxn];
bool vis[maxn];
int fr[maxn];
int m, n, k;
int cnt;
int id[maxn][maxn];

struct edge
{
int to, next;
} e[maxn];
int head[maxn];
int index;

void addedge(int u, int v)
{
e[index].to = v;
e[index].next = head[u];
head[u] = index;
index++;
}

bool match(int v)
{
for (int i = head[v]; i != -1; i = e[i].next)
{
int u = e[i].to;
if (!vis[u])
{
vis[u] = true;
if (fr[u] == -1 || match(fr[u]))
{
fr[u] = v;
return true;
}
}
}
return false;
}

int hungary()
{
int ret = 0;
memset(fr, -1, sizeof(fr));
for (int i = 1; i <= cnt; i++)
{
memset(vis, false, sizeof(vis));
if (match(i))
ret++;
}
return ret;
}

int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif // LOCAL
while (~scanf("%d%d%d", &m, &n, &k))
{
memset(maze, true, sizeof(maze));
for (int i = 0; i < k; i++)
{
int x, y;
scanf("%d%d", &y, &x);
maze[x][y] = false;
}
if ((m * n - k) % 2)
{
printf("NO\n");
}
else
{
cnt = 0;
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
if (maze[i][j])
{
id[i][j] = ++cnt;
}
}
}
memset(head, -1, sizeof(head));
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
if (maze[i][j])
{
for (int d = 0; d < 4; d++)
{
int ni = i + dir[d][0];
int nj = j + dir[d][1];
if (1 <= ni && ni <= m && 1 <= nj && nj <= n && maze[ni][nj])
{
addedge(id[i][j], id[ni][nj]);
}
}
}
}
}
//cout << hungary() <<endl;
if (hungary() + k == m * n)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
}
}
return 0;
}

poj 1469：

题意：

给课程和喜欢这门课的学生。

问是否存在每个不同的学生喜欢不同的课。

解析：

最直接的二分图匹配。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <climits>
#include <cassert>
#define LL long long
#define lson lo, mi, rt << 1
#define rson mi + 1, hi, rt << 1 | 1

using namespace std;
const int maxn = 300 + 10;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double ee = exp(1.0);

vector<int> g[maxn];
int fr[maxn];
bool vis[maxn];
int p, n;

bool match(int v)
{
for (int i = 0; i < g[v].size(); i++)
{
int u = g[v][i];
if (!vis[u])
{
vis[u] = true;
if (fr[u] == -1 || match(fr[u]))
{
fr[u] = v;
return true;
}
}
}
return false;
}

int angary()
{
int ret = 0;
memset(fr, -1, sizeof(fr));
for (int i = 1; i <= p; i++)
{
memset(vis, false, sizeof(vis));
if (match(i))
{
ret++;
}
}
return ret;
}

int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif // LOCAL
int ncase;
scanf("%d", &ncase);
while (ncase--)
{
for (int i = 0; i < maxn; i++)
g[i].clear();
scanf("%d%d", &p, &n);
for (int i = 1; i <= p; i++)
{
int k;
scanf("%d", &k);
while (k--)
{
int x;
scanf("%d", &x);
g[i].push_back(x);
}
}
//    cout << angary() << endl;
if (angary() == p)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}

poj 1274

题意：

二分图最大匹配。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <climits>
#include <cassert>
#define LL long long
#define lson lo, mi, rt << 1
#define rson mi + 1, hi, rt << 1 | 1

using namespace std;
const int maxn = 200 + 10;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double ee = exp(1.0);

vector<int> g[maxn];
int fr[maxn];
bool vis[maxn];
int n, m;

bool match(int v)
{
for (int i = 0; i < g[v].size(); i++)
{
int u = g[v][i];
if (!vis[u])
{
vis[u] = true;
if (fr[u] == -1 || match(fr[u]))
{
fr[u] = v;
return true;
}
}
}
return false;
}

int hungary()
{
int ret = 0;
memset(fr, -1, sizeof(fr));
for (int i = 1; i <= n; i++)
{
memset(vis, false, sizeof(vis));
if (match(i))
{
ret++;
}
}
return ret;
}

int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif // LOCAL
while (~scanf("%d%d", &n, &m))
{
for (int i = 0; i <= n; i++)
g[i].clear();
for (int i = 1; i <= n; i++)
{
int k;
scanf("%d", &k);
while (k--)
{
int x;
scanf("%d", &x);
g[i].push_back(x);
}
}
printf("%d\n", hungary());
}
return 0;
}