hdu3360-二分图

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3360

Description
略略略……

Input

Your program will be tested on one or more test cases. Each test case is specified using R+1 lines.
The first line specifies two integers (1<= R,C <= 50) which are the dimensions of the museum hall. The next R lines contain C integers separated by one or more spaces. The j-th integer of the i-th row is -1 if cell (i, j) already contains one of the museum’s guards, otherwise it contains an integer (0 <= T <= 2 12) representing the type of the artifact in that cell.
The last line of the input file has two zeros.

Output
For each test case, print the following line:
k. G
Where k is the test case number (starting at one,) and G is the minimum number of additional guards to hire such that all remaining artifacts are secured.

Sample Input
1 3
512 -1 2048
2 3
512 2560 2048
512 2560 2048
0 0

Sample Output

1. 0
2. 2

题意
博物馆大小为n*m，博物馆不为-1的点上有文物，要求所有文物被保护需要的最少护卫。护卫的位置不能在博物馆之外。
文物被保护的定义：这个文物的所有必须被保护的点都要有守卫。 一个文物由一个2的12次方以内的数表示，那么这个数最多有12位，对应图中给出打的12个可能要被保护的点。

思路
把每个文物与之需要保护的点建边，要求所有的边都被覆盖，就是求二分图的最小点覆盖。

---

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81

#include #include #include #include #include #include #include #include #include #include #include #define lson l,mid,i<<1 #define rson mid+1,r,i<<1|1 using namespace std; typedef long long ll; const int maxn = 3600; int dx[12] = { -1,-2,-2,-1,1,2,2,1,-1,0,1,0 }; int dy[12] = { -2,-1,1,2,2,1,-1,-2,0,1,0,-1 }; int link[maxn]; int vis[maxn]; vector<int>mp[maxn]; int a[maxn][maxn]; int Nx; bool dfs(int u) { for (int i = 0; i < mp[u].size(); i++) { int v = mp[u][i]; if (!vis[v]) { vis[v] = 1; if (link[v] == -1 || dfs(link[v])) { link[v] = u; return 1; } } } return 0; } int xyl() { int ans = 0; memset(link, -1, sizeof(link)); for (int i = 0; i < Nx; i++) { memset(vis, 0, sizeof(vis)); if (dfs(i)) ans++; } return ans; } int main() { int n, m; int cnt = 1; while (scanf("%d%d", &n, &m) != EOF) { if (n == 0 && m == 0) break; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { scanf("%d", &a[i][j]); } } for (int i = 0; i < n*m; i++) mp[i].clear(); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { if (a[i][j] != -1) { for (int k = 0; k < 12; k++) { if ((a[i][j] >> k) & 1) { int xx = dx[k] + i; int yy = dy[k] + j; if (xx >= 0 && xx < n&&yy >= 0 && yy < m&&a[xx][yy] != -1) { int u = i * m + j; int v = xx * m + yy; mp[u].push_back(v); mp[v].push_back(u); } } } } } } Nx = n * m; printf("%d. %d\n", cnt++, xyl() / 2); } }

赏

谢谢你请我吃糖果