POJ 2446 Chessboard（二分图匹配）

POJ 2446 Chessboard

题目链接

题意：给定一个棋盘，上面有洞，问能否完美用1 * 2纸条铺满，纸条不能重叠，不能贴到洞上

思路：二分图匹配，求最大独立集，相邻的空白建边即可，然后这题有一个很坑的地方啊，就是输入洞的位置(x, y)居然是先输y再输x，这不合常理坑爹

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;

const int N = 35;
const int d[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};

int n, m, k, graph

;
vector<int> g[N * N];
int left[N * N], vis[N * N];

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

int hungary() {
int ans = 0;
memset(left, -1, sizeof(left));
for (int i = 0; i < n * m; i++) {
memset(vis, 0, sizeof(vis));
if (dfs(i)) ans++;
}
return ans;
}

int main() {
while (~scanf("%d%d%d", &n, &m, &k)) {
int x, y;
memset(graph, 0, sizeof(graph));
for (int i = 0; i < k; i++) {
scanf("%d%d", &y, &x);
x--; y--;
graph[x][y] = 1;
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
g[i * m + j].clear();
if (graph[i][j]) continue;
for (int k = 0; k < 4; k++) {
int x = i + d[k][0];
int y = j + d[k][1];
if (x < 0 || x >= n || y < 0 || y >= m || graph[x][y]) continue;
g[i * m + j].push_back(x * m + y);
}
}
}
printf("%s\n", n * m - k == hungary() ? "YES" : "NO");
}
return 0;
}