1034.Head of a Gang (30)

1034.Head of a Gang (30)

pat-al-1034

2017-02-04

• 图的遍历，连通分量，dfs
• 对于每一个连通分量，要求其head和包含的成员个数，以及该gang的总权值
• 套路：用map来映射名称和序号、visited、边e、节点权重weight
• 练习了用迭代器输出map的内容
• 坑见注释

/** * pat-al-1034 * 2017-02-04 * Cpp version * Author: fengLian_s */#include<stdio.h>#include<iostream>#include<map>#include<string>#define MAX 2003//坑……我还以为是1000，导致段错误using namespace std;int k;int maxNumberOfPeople = 1;int visited[MAX], weight[MAX], e[MAX][MAX];map<string, int> str2int;map<int, string> int2str;map<string, int> result;int string2int(string s){  if(str2int[s] == 0)//还未放入映射中  {    str2int[s] = maxNumberOfPeople;    int2str[maxNumberOfPeople] = s;    //cout << s << ": " << maxNumberOfPeople << endl;    return maxNumberOfPeople++;  }  else    return str2int[s];}void dfs(int u, int &head, int &numberOfMember, int &totalWeight){  visited[u] = 1;  numberOfMember++;  if(weight[u] > weight[head])    head = u;  for(int v = 1;v < maxNumberOfPeople;v++)  {    if(e[u][v] != 0)//坑：和下面的visited必须要分开判断，此处判断的是是否有边没有加入总权重    {      totalWeight += e[u][v];      //printf("e[%d][%d] = %d\n", u, v, e[u][v]);      e[u][v] = e[v][u] = 0;//坑，一条路只能加一遍      if(visited[v] == 0)//坑：visited判断的时候是否继续深搜        dfs(v, head, numberOfMember, totalWeight);    }  }}int main(){  int n;  scanf("%d%d", &n, &k);  for(int i = 0;i < n;i++)  {    string s1, s2;    int w;    cin >> s1 >> s2 >> w;    int id1, id2;    id1 = string2int(s1);    id2 = string2int(s2);    weight[id1] += w;    weight[id2] += w;    e[id1][id2] += w;    e[id2][id1] += w;//坑，不用+=的话会被新的权值覆盖  }  //solve:  for(int i = 1;i < maxNumberOfPeople;i++)  {    if(visited[i] == 0)    {      int head = i, totalWeight = 0, numberOfMember = 0;      dfs(i, head, numberOfMember, totalWeight);      //cout << "head = " << head << ", numberOfMember = " << numberOfMember << ", totalWeight = " << totalWeight << endl;      if(numberOfMember > 2 && totalWeight > k)        result[int2str[head]] = numberOfMember;    }  }  //output:  printf("%lu\n", result.size());  for(map<string, int>::iterator it = result.begin();it != result.end();it++)    cout << it->first << " " << it->second << endl;  return 0;}

-FIN-