【牛客网】KY118 继续畅通工程

这道题可以使用并查集解决，利用Kruskal算法的思想构建最小生成树，其中，我使用优先队列根据修建状态进行排序，优先已建的路：

#include<iostream>
#include<string>
#include<queue>

using namespace std;

const int N = 100 + 10;

int arr[N];

struct Distance{
    int n1, n2, dis, state;
    Distance(int n1, int n2, int d, int s): n1(n1), n2(n2), dis(d), state(s){}
    bool operator>(Distance d) const{
        if(state == d.state) return dis > d.dis;
        return state < d.state;
    }
};

void Initial(int *arr, int n){
    for(int i = 1; i <= n; i++) arr[i] = -1;
}

int Find(int *arr, int x){
    int root = x;
    while(arr[root] > 0) root = arr[root];
    while(x != root){
        int t = arr[x];
        arr[x] = root;
        x = t;
    }
    return root;
}

bool Union(int *arr, int x, int y){
    int root1 = Find(arr, x);
    int root2 = Find(arr, y);
    if(root1 != root2){
        if(arr[root2] > arr[root1]){
            arr[root1] += arr[root2];
            arr[root2] = root1;
        }else{
            arr[root2] += arr[root1];
            arr[root1] = root2;
        }
        return true;
    }
    return false;
}

int main(){
    int n, a, b, dis, state;
    while(scanf("%d", &n) != EOF){
        if(n == 0) break;
        int len = (n - 1) * n / 2;
        priority_queue<Distance, vector<Distance>, greater<Distance>> pq;
        Initial(arr, n);
        for(int i = 0; i < len; i++){
            scanf("%d%d%d%d", &a, &b, &dis, &state);
            pq.push(Distance(a, b, dis, state));
        }
        len = pq.size();
        int sum = 0;
        for(int i = 0; i < len; i++){
            Distance d = pq.top();
            pq.pop();
            if(Union(arr, d.n1, d.n2)){
                if(!d.state) sum += d.dis;
            }
        }
        printf("%d\n", sum);
    }
    return 0;
}

标签：arr,int,牛客,state,KY118,畅通,root2,root1,dis
来源： https://blog.csdn.net/weixin_55267022/article/details/122799525