代码随想录算法训练营Day-55 图论06 | 108.冗余连接、109.冗余连接II

108.冗余连接

并查集判断冗余边：边的两个端点已经连通，则这条边就是成环边；题目要最后出现的成环边，所以遍历完所有边，输出最后记录的一条。

cpp 复制代码

#include<iostream>
#include<vector>
using namespace std;

int n=1005;
vector<int> father = vector<int>(n,0);

void _init(){
    for(int i=0;i<n;i++){
        father[i] = i;
    }
}

int find(int a){
    if(father[a] == a) return a;
    else return father[a] = find(father[a]);
}

void join(int a,int b){
    a = find(a);
    b = find(b);
    if(a==b) return;
    else father[a] = b;
}

bool isSame(int a,int b){
    a = find(a);
    b = find(b);
    return a==b;
}

int main(){
    int s,t;
    cin>>n;
    _init();
    int ans_s=0,ans_t=0;
    for(int i=0;i<n;i++){
        cin>>s>>t;
        if(isSame(s,t)){
            ans_s = s;
            ans_t = t;
        }else join(s,t);
    }
    cout<<ans_s<<" "<<ans_t<<endl;
}

109.冗余连接II

两种情况：

1.存在入度为2的节点-找到连到该节点的两条边，优先删除后出现的边，如果删除后无环，就直接输出该边，否则删除前面那条。

2.不存在入度为2的节点，说明有环，退化成上一题并查集寻找冗余边。

cpp 复制代码

#include<iostream>
#include<vector>
using namespace std;

int n=1005;
vector<int> father = vector<int>(n,0);

void _init(){
    for(int i=0;i<n;i++){
        father[i] = i;
    }
}

int find(int a){
    if(father[a] == a) return a;
    else return father[a] = find(father[a]);
}

void join(int a,int b){
    a = find(a);
    b = find(b);
    if(a==b) return;
    else father[a] = b;
}

bool isSame(int a,int b){
    a = find(a);
    b = find(b);
    return a==b;
}

bool isTreeAfterRemoved(vector<vector<int>>& edges, int remove){
    for(int i=0;i<edges.size();i++){
        if(i == remove) continue;

        if(isSame(edges[i][0],edges[i][1])) return false;
        else join(edges[i][0],edges[i][1]);
    }
    return true;
}

int main(){
    _init();
    cin>>n;
    
    vector<vector<int>> edges(n,vector<int>(2));
    vector<int> indegree(n+1, 0);

    //记录所有有向边及顺序，并统计每个节点的入度
    for(int i=0;i<n;i++){
        cin>>edges[i][0]>>edges[i][1];
        indegree[edges[i][1]]++;
    }

    int twoIndegreedot=0;
    for(int i=1;i<=n;i++){
        if(indegree[i] == 2) twoIndegreedot = i;
    }
    if(twoIndegreedot!=0){
        int edge1 = -1, edge2 = -1;
        for(int i=0;i<edges.size();i++){
            if(edges[i][1] == twoIndegreedot){
                if(edge1==-1) edge1 = i;
                else edge2 = i;
            }
        }
        if(isTreeAfterRemoved(edges, edge2)){
            cout<<edges[edge2][0]<<" "<<edges[edge2][1];
            return 0;
        }else{
            cout<<edges[edge1][0]<<" "<<edges[edge1][1];
            return 0;
        }
    }else{//否则就只是去掉最后一个成环边就行
        int ans_1=0,ans_2=0;
        for(int i=0;i<edges.size();i++){
            if(isSame(edges[i][0],edges[i][1])){
                ans_1 = edges[i][0];
                ans_2 = edges[i][1];
            }else join(edges[i][0],edges[i][1]);
        }
        cout<<ans_1<<" "<<ans_2<<endl;
        return 0;
    }


}