PAT_A1034#Head of a Gang

Source:

PAT A1034 Head of a Gang (30 分)

Description:

One way that the police finds the head of a gang is to check people's phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of all the phone calls made between the two persons. A "Gang" is a cluster of more than 2 persons who are related to each other with total relation weight being greater than a given threthold K. In each gang, the one with maximum total weight is the head. Now given a list of phone calls, you are supposed to find the gangs and the heads.

Input Specification:

Each input file contains one test case. For each case, the first line contains two positive numbers Nand K (both less than or equal to 1000), the number of phone calls and the weight threthold, respectively. Then N lines follow, each in the following format:

Name1 Name2 Time

where Name1 and Name2 are the names of people at the two ends of the call, and Time is the length of the call. A name is a string of three capital letters chosen from A-Z. A time length is a positive integer which is no more than 1000 minutes.

Output Specification:

For each test case, first print in a line the total number of gangs. Then for each gang, print in a line the name of the head and the total number of the members. It is guaranteed that the head is unique for each gang. The output must be sorted according to the alphabetical order of the names of the heads.

Sample Input 1:

8 59
AAA BBB 10
BBB AAA 20
AAA CCC 40
DDD EEE 5
EEE DDD 70
FFF GGG 30
GGG HHH 20
HHH FFF 10

Sample Output 1:

2
AAA 3
GGG 3

Sample Input 2:

8 70
AAA BBB 10
BBB AAA 20
AAA CCC 40
DDD EEE 5
EEE DDD 70
FFF GGG 30
GGG HHH 20
HHH FFF 10

Sample Output 2:

0

Keys:

Attention:

  • 给出N条边，最多可能有2*N个结点；
  • 边权累加的操作如果放在 if 里面的话，图中如果回路，会少数一条边；
  • 边权累加的操作如果放在 if 外面的话，相当于各个边数了两次，结果除以2就可以了；
  • 边权累加的操作正常应该放在 if(grap[u][v]!=INF)的里面，不过这题边权初始化为0，加上无效边不影响结果；
  • 输入的时候预处理各个结点的边权之和，是个小的tips，可以注意一下；

Code:

  1 /*
  2 Data: 2019-04-14 19:46:23
  3 Problem: PAT_A1034#Head of a Gang
  4 AC: 61:22
  5 
  6 题目大意：
  7 A和B有通话总时长表示他们的联系程度，
  8 如果一撮人通话总时长超过某一阈值，则认定为“犯罪团伙”，其中电话打的最多的就是头目；
  9 输入：
 10 第一行给出通话数目N和阈值K（<=1e3）
 11 接下来N行给出 v1 v2 w，其中V用三个大写字母表示，w<=1e3
 12 输出：
 13 第一行输出团伙数目
 14 接下来按照字典序依次输出各个团伙的头目及其人数
 15 
 16 基本思路：
 17 预处理各顶点边权之和，
 18 遍历各个连通分量，统计结点数目，边权之和，犯罪头目
 19 map存储犯罪头目及其人数，达到字典序排序的目的
 20 */
 21 #include<cstdio>
 22 #include<iostream>
 23 #include<algorithm>
 24 #include<string>
 25 #include<map>
 26 using namespace std;
 27 const int M=2e3+10;
 28 int grap[M][M],vis[M],cost[M],pt=1;
 29 int K,k,sum,Max;
 30 map<string,int> mp,gang;
 31 string toS[M],head;
 32 
 33 int ToN(string s)
 34 {
 35     if(mp[s]==0)
 36     {
 37         mp[s]=pt;
 38         toS[pt]=s;
 39         return pt++;
 40     }
 41     else
 42         return mp[s];
 43 }
 44 
 45 void DFS(int u)
 46 {
 47     vis[u]=1;
 48     sum++;
 49     if(cost[u] > Max)
 50     {
 51         Max = cost[u];
 52         head = toS[u];
 53     }
 54     for(int v=1; v<pt; v++)
 55     {
 56         k += grap[u][v];
 57         if(grap[u][v]!=0 && vis[v]==0)
 58                 DFS(v);
 59         /*若按规矩办事
 60         if(grap[u][v]!=0)           //判断是否存在边
 61         {
 62             k += grap[u][v];        //各边数了两次
 63             if(vis[v]==0)
 64                 k += grap[u][v];    //若有回路，少数一条边
 65         }
 66         */
 67     }
 68 }
 69 
 70 int Travel()
 71 {
 72     fill(vis,vis+M,0);
 73     int cnt=0;
 74     for(int v=1; v<pt; v++)
 75     {
 76         k=0;sum=0;Max=0;
 77         if(vis[v]==0)
 78         {
 79             DFS(v);
 80             if(k>K*2 && sum>2)
 81             {
 82                 cnt++;
 83                 gang[head]=sum;
 84             }
 85         }
 86     }
 87     return cnt;
 88 }
 89 
 90 int main()
 91 {
 92 #ifdef  ONLINE_JUDGE
 93 #else
 94     freopen("Test.txt", "r", stdin);
 95 #endif // ONLINE_JUDGE
 96 
 97     int n;
 98     scanf("%d%d", &n,&K);
 99     fill(grap[0],grap[0]+M*M,0);
100     fill(cost,cost+M,0);
101     for(int i=0; i<n; i++)
102     {
103         string s1,s2;
104         int v1,v2,w;
105         cin >> s1 >> s2 >> w;
106         v1 = ToN(s1);
107         v2 = ToN(s2);
108         grap[v1][v2]+=w;
109         grap[v2][v1]+=w;
110         cost[v1] += w;
111         cost[v2] += w;
112     }
113     int cnt=Travel();
114     printf("%d\n", cnt);
115     for(auto it=gang.begin(); it!=gang.end(); it++)
116         cout << it->first << " " << it->second << endl;
117 
118     return 0;
119 }

 

posted @ 2019-05-18 22:54  林東雨  阅读(207)  评论(0)  编辑  收藏  举报