算法笔记--图

题目：PAT A1034

 

题解：

1. 姓名---编号 stringtoint map<string,int>,inttostring map<int,string>
2. 每个人的点权可以在读入的时候直接处理==边权之和
3. 图的遍历分为两步 ：：：：先遍历联通快，在便利图
4. 注意联通环  需要在递归进行遍历的时候先进性边权相加，在进行递归访问节点的问题

#include <iostream>
#include <map>
#include <string>

using namespace std;

const int maxn = 2100;
const int INF = 10000000;

map<int, string> inttostring;
map<string, int> stringtoint;
map<string, int> Gang;

int G[maxn][maxn] = { 0 }, weight[maxn] = { 0 };
int n, k, numperson=0;
bool vis[maxn] = { false };
//图有环，先累加编圈，在遍历节点,单个连通图的遍历
void dfs(int nowlist, int &head,int &number,int &totalvalue)
{
    number++;
    vis[nowlist] = true;
    if (weight[nowlist] > weight[head])
    {
        head = nowlist;
    }
    for (int i = 0; i < numperson; ++i)
    {
        if (G[nowlist][i] > 0)
        {
            totalvalue += G[nowlist][i];
            G[nowlist][i] = G[i][nowlist] = 0;
            if (vis[i] == false)
            {
                dfs(i, head, number, totalvalue);
            }
        }
    }
}

//遍历整个图 获取每个联通快的信息

void dfstrave() {
    for (int i = 0; i < numperson; ++i)
    {
        if (vis[i] == false)
        {
            int head = i, num = 0, tot = 0;
                dfs(i, head, num, tot);
                if (num > 2 && tot > k)
                {
                    Gang[inttostring[head]] = num;
                }
        }
    }
}

int change(string str)
{
    if (stringtoint.find(str) != stringtoint.end())
        return stringtoint[str];
    else {
        stringtoint[str] = numperson;
        inttostring[numperson] = str;
        return numperson++;
    }
}

int main()
{
    int w;
    string str1, str2;
    cin >> n >> k;
    for (int i = 0; i < n; ++i)
    {
        cin >> str1 >> str2 >> w;
        int id1 = change(str1);
        int id2 = change(str2);
        weight[id1] += w;
        weight[id2] += w;
        G[id1][id2] += w;
        G[id2][id1] += w;
    }
    dfstrave();
    cout << Gang.size() << endl;
    for (auto i : Gang)
        cout << i.first << " " << i.second << endl;
    system("pause");
    return 0;
}