1034 Head of a Gang (30分)

Posted 2020-11-27 d-i-p

1. 题目

2. 思路

使用并查集归属集合

3. 注意点

1. 可以使用map<string, string> 模拟int类型的并查集，减少了string和int类型转换的问题
   因为c++的map，值如果不存在会自动初始化

map<string, string> father; //定义

// 查找root
string findfather(string x){
    if(father[x] == ""){
        father[x] = x;
    }else{
        while(father[x] != x){
            x = father[x];
        }
    }
    return x;
}

1. 对于集合的总权重的计算， 存放map<string, int> sum

• 如果两个name本来就是一个节点，sum[findfather(n1)] + weight;
• 如果两个name不是一个节点， sum[findfather(n1)] + sum[findfather(n2)] + weight

4. 代码

#include<cstdio>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<string>

using namespace std;

map<string, string> father;
map<string, int> weights;
map<string, int> sum;

string findfather(string x){
    if(father[x] == ""){
        father[x] = x;
    }else{
        while(father[x] != x){
            x = father[x];
        }
    }
    return x;
}

bool cmp(string s1, string s2){
    return weights[s1] > weights[s2];
}

void collect(int k){
    vector<string> res;
    map<string, set<string> > col;
    for(auto a:father){
        col[findfather(a.first)].insert(a.first);
    }
    for(auto a:col){
        if(a.second.size() > 2 && sum[findfather(a.first)] > k){
            vector<string> temp(a.second.begin(), a.second.end());
            sort(temp.begin(), temp.end(), cmp);
            res.push_back(temp[0]);     
        }
    }
    sort(res.begin(), res.end(), [](string s1, string s2){
        return s1 < s2;
    });
    printf("%d
", res.size());
    for(auto a:res){
        printf("%s %d
", a.c_str(), col[findfather(a)].size());
    }
}

int main(){
    int n, k;
    scanf("%d %d", &n, &k);
    char n1[10], n2[10];
    int w;
    for(int i=0;i<n;i++){
        scanf("%s %s %d", n1, n2, &w);
        weights[n1] += w;
        weights[n2] += w;
        string f1 = findfather(n1);
        string f2 = findfather(n2);
        if(f1 != f2){
            sum[f1] += sum[f2];
            father[f2] = f1;
        }
        sum[f1] += w; 
    }
    collect(k);
}