1034.Head of a Gang (30)

2019-11-14 10:14:50

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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;}

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