题目链接
题意:给定一个棋盘,上面有黑白或者空,现在有一种L型拼图如图,问能否拼出给定图案,拼图不能重叠
思路:二分图匹配,拆点,对于每个黑点,拆点两个点,一个和横向连,一个和纵向连,然后进行二分图最大匹配,如果能完美匹配就是正确的
代码:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 250005; const int MAXN = 1000005; struct Graph { struct Edge { int u, v, val, next; Edge() {} Edge(int u, int v, int val = 0) { this->u = u; this->v = v; this->val = val; } } edge[MAXN]; int first[N], en; void init() { memset(first, -1, sizeof(first)); en = 0; } void add(int u, int v, int val = 0) { edge[en] = Edge(u, v, val); edge[en].next = first[u]; first[u] = en++; } } g; const int d[4][2] = {{0, -1}, {-1, 0}, {1, 0}, {0, 1}}; int t, n, m, b, w, g2[505][505], match[N], vis[N]; char str[505][505]; bool dfs(int u) { for (int i = g.first[u]; i != -1; i = g.edge[i].next) { int v = g.edge[i].v; if (vis[v]) continue; vis[v] = 1; if (match[v] == -1 || dfs(match[v])) { match[v] = u; return true; } } return false; } bool hungary() { int sum = 0; memset(match, -1, sizeof(match)); for (int i = 0; i < 2 * b; i++) { memset(vis, 0, sizeof(vis)); if (!dfs(i)) return false; } return true; } int main() { scanf("%d", &t); while (t--) { scanf("%d%d", &n, &m); b = 0; w = 0; g.init(); for (int i = 0; i < n; i++) { scanf("%s", str[i]); for (int j = 0; j < m; j++) { if (str[i][j] == 'B') g2[i][j] = b++; else if (str[i][j] == 'W') g2[i][j] = w++; } } if (b * 2 != w) { printf("NO\n"); continue; } for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (str[i][j] == 'B') { 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 || str[x][y] != 'W') continue; if (y == j) g.add(g2[i][j] + b, g2[x][y]); else g.add(g2[i][j], g2[x][y]); } } } } if (hungary()) printf("YES\n"); else printf("NO\n"); } return 0; }