Bus Pass BFS搜索

  1 /********************************
  2 please don‘t hack me!! /(ToT)/~~
  3                 __------__
  4               /~          ~  5              |    //^\\//^\|
  6            /~~\  ||  T| |T|:~  7           | |6   ||___|_|_||:|
  8            \__.  /      o  \/‘
  9             |   (       O   )
 10    /~~~~\    `\  \         /
 11   | |~~\ |     )  ~------~` 12  /‘ |  | |   /     ____ /~~~) 13 (_/‘   | | |     /‘    |    ( |
 14        | | |     \    /   __)/  15        \  \ \      \/    /‘ \   ` 16          \  \|\        /   | |\___|
 17            \ |  \____/     | |
 18            /^~>  \        _/ <
 19           |  |         \        20           |  | \        \         21           -^-\  \       |        )
 22                `\_______/^\______/
 23 ************************************/
 24 
 25 #include <iostream>
 26 #include <cstdio>
 27 #include <cstring>
 28 #include <cmath>
 29 #include <algorithm>
 30 #include <string>
 31 #include <vector>
 32 #include <set>
 33 #include <map>
 34 #include <queue>
 35 #include <stack>
 36 #include <cstdlib>
 37 #include <sstream>
 38 using namespace std;
 39 typedef long long LL;
 40 const LL INF = 0x5fffffff;
 41 const double EXP = 1E-8;
 42 const LL MOD = (LL)1E9+7;
 43 const int MS = 10000;
 44 
 45 int nz, nr;
 46 int cnt[MS];
 47 int edges[MS][11];
 48 int dis[MS];
 49 int reach[MS];   //可以bfs防止重复访问某节点，也可以判断是否存在标号为i的地区
 50 int cur;         //reach[i] == cur. 表示地区i在第cur+ 1站访问到。
 51 
 52 void bfs(int x) {
 53     int val = 1;  // 层次
 54     queue<int> que[2];   //  层次扩展的滚动队列
 55     int a = 0, b = 1;
 56     if (reach[x] < cur) {
 57         reach[x] = cur;
 58         que[b].push(x);
 59         dis[x] = max(dis[x],val);
 60         val++;
 61     }
 62     while (!que[b].empty()) {
 63         swap(a,b);
 64         while (!que[a].empty()) {
 65             int u = que[a].front();
 66             que[a].pop();
 67             for (int i = 0; i < cnt[u]; i++) {
 68                 int v = edges[u][i];
 69                 if (reach[v] < cur) { //如果在本站中访问了，那么reach[v] == cur.
 70                     reach[v] = cur;
 71                     dis[v] = max(dis[v],val);
 72                     que[b].push(v);
 73                 }
 74             }
 75         }
 76         val++;
 77     }
 78 }
 79 
 80 int main() {
 81     int T;
 82     scanf("%d",&T);
 83     while (T--) {
 84         scanf("%d%d",&nz,&nr);
 85         int id,x;
 86         for (int i = 0; i < nz; i++) {
 87             scanf("%d",&id);
 88             scanf("%d",&cnt[id]);
 89             for (int j = 0; j < cnt[id]; j++) {
 90                 scanf("%d",&edges[id][j]);
 91             }
 92         }
 93         memset(dis,0,sizeof(dis));
 94         memset(reach,-1,sizeof(reach));
 95         cur = 0;
 96         for (int i = 0; i < nr; i++) {
 97             int n;
 98             scanf("%d",&n);
 99             for (int j = 0; j < n; j++) {
100                 scanf("%d",&x);
101                 bfs(x);
102                 cur++;
103             }
104         }
105         int minv = INF;
106         int center = 0;
107         for (int i = 0; i < MS; i++) {
108             //因为题目中并没有说编号是连续的，地区的编号可能是随意的。
109             //我们可以用reach[i] == cur -1 来判断是否存在编号为i的地区。
110             if (reach[i] == cur -1 && dis[i] < minv) {
111                 minv = dis[i];
112                 center = i;
113             }
114         }
115         printf("%d %d\n",minv,center);
116     }
117     return 0;
118 }