Separate Connections
Time Limit:5000MS Memory Limit:65536KB
Total Submit:421 Accepted:41
Description
Partychen are analyzing a communications network with at most 18 nodes. Character in a matrix i,j (i,j both 0-based,as matrix[i][j]) denotes whether nodes i and j can communicate ('Y' for yes, 'N' for no). Assuming a node cannot communicate with two nodes at once, return the maximum number of nodes that can communicate simultaneously. If node i is communicating with node j then node j is communicating with node i.
Input
The first line of input gives the number of cases, N(1 ≤ N ≤ 100). N test cases follow.
In each test case,the first line is the number of nodes M(1 ≤ M ≤ 18),then there are a grid by M*M describled the matrix.
Output
For each test case , output the maximum number of nodes that can communicate simultaneously
Sample Input
2
5
NYYYY
YNNNN
YNNNN
YNNNN
YNNNN
5
NYYYY
YNNNN
YNNNY
YNNNY
YNYYN
Sample Output
2
4
Hint
The first test case:
All communications must occur with node 0. Since node 0 can only communicate with 1 node at a time, the output value is 2.
The second test case:
In this setup, we can let node 0 communicate with node 1, and node 3 communicate with node 4.
Source
解题：逼格很高啊！带花树模板题，可惜本人不会敲，只好找了份模板
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <queue> 5 using namespace std; 6 const int N = 250; 7 int belong[N]; 8 int findb(int x) { 9 return belong[x] == x ? x : belong[x] = findb(belong[x]); 10 } 11 void unit(int a, int b) { 12 a = findb(a); 13 b = findb(b); 14 if (a != b) belong[a] = b; 15 } 16 17 int n, match[N]; 18 vector<int> e[N]; 19 int Q[N], rear; 20 int nxt[N], mark[N], vis[N]; 21 int LCA(int x, int y) { 22 static int t = 0; 23 t++; 24 while (true) { 25 if (x != -1) { 26 x = findb(x); 27 if (vis[x] == t) return x; 28 vis[x] = t; 29 if (match[x] != -1) x = nxt[match[x]]; 30 else x = -1; 31 } 32 swap(x, y); 33 } 34 } 35 36 void group(int a, int p) { 37 while (a != p) { 38 int b = match[a], c = nxt[b]; 39 if (findb(c) != p) nxt[c] = b; 40 if (mark[b] == 2) mark[Q[rear++] = b] = 1; 41 if (mark[c] == 2) mark[Q[rear++] = c] = 1; 42 43 unit(a, b); 44 unit(b, c); 45 a = c; 46 } 47 } 48 49 // 增广 50 void aug(int s) { 51 for (int i = 0; i < n; i++) // 每个阶段都要重新标记 52 nxt[i] = -1, belong[i] = i, mark[i] = 0, vis[i] = -1; 53 mark[s] = 1; 54 Q[0] = s; 55 rear = 1; 56 for (int front = 0; match[s] == -1 && front < rear; front++) { 57 int x = Q[front]; // 队列Q中的点都是S型的 58 for (int i = 0; i < (int)e[x].size(); i++) { 59 int y = e[x][i]; 60 if (match[x] == y) continue; // x与y已匹配，忽略 61 if (findb(x) == findb(y)) continue; // x与y同在一朵花，忽略 62 if (mark[y] == 2) continue; // y是T型点，忽略 63 if (mark[y] == 1) { // y是S型点，奇环缩点 64 int r = LCA(x, y); // r为从i和j到s的路径上的第一个公共节点 65 if (findb(x) != r) nxt[x] = y; // r和x不在同一个花朵，nxt标记花朵内路径 66 if (findb(y) != r) nxt[y] = x; // r和y不在同一个花朵，nxt标记花朵内路径 67 68 // 将整个r -- x - y --- r的奇环缩成点，r作为这个环的标记节点，相当于论文中的超级节点 69 group(x, r); // 缩路径r --- x为点 70 group(y, r); // 缩路径r --- y为点 71 } else if (match[y] == -1) { // y自由，可以增广，R12规则处理 72 nxt[y] = x; 73 for (int u = y; u != -1; ) { // 交叉链取反 74 int v = nxt[u]; 75 int mv = match[v]; 76 match[v] = u, match[u] = v; 77 u = mv; 78 } 79 break; // 搜索成功，退出循环将进入下一阶段 80 } else { // 当前搜索的交叉链+y+match[y]形成新的交叉链，将match[y]加入队列作为待搜节点 81 nxt[y] = x; 82 mark[Q[rear++] = match[y]] = 1; // match[y]也是S型的 83 mark[y] = 2; // y标记成T型 84 } 85 } 86 } 87 } 88 89 bool g[N][N]; 90 char mp[N][N]; 91 int main() { 92 int kase; 93 scanf("%d",&kase); 94 while(kase--) { 95 scanf("%d", &n); 96 for (int i = 0; i < n; i++) 97 for (int j = 0; j < n; j++) g[i][j] = false; 98 for(int i = 0; i < n; ++i) e[i].clear(); 99 for(int i = 0; i < n; ++i){ 100 scanf("%s",mp[i]); 101 for(int j = 0; j < n; ++j){ 102 if(mp[i][j] == 'Y'){ 103 e[i].push_back(j); 104 e[j].push_back(i); 105 g[i][j] = g[j][i] = true; 106 } 107 } 108 } 109 for (int i = 0; i < n; i++) match[i] = -1; 110 for (int i = 0; i < n; i++) if (match[i] == -1) aug(i); 111 int tot = 0; 112 for (int i = 0; i < n; i++) if (match[i] != -1) tot++; 113 printf("%d ", tot); 114 } 115 return 0; 116 }