ECNUOJ 2575 Separate Connections

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\n", tot);
114     }
115     return 0;
116 }
View Code

 

posted @ 2015-07-08 15:52  狂徒归来  阅读(475)  评论(0)  编辑  收藏  举报