浙大上机题--继续畅通工程

题目链接

要先计算连通分支的个数，最终只有一个连通分支即可

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 5010;

struct node {
    int wei, x, y, sta;
    bool operator < (const node& t) const {
        return wei < t.wei;
    }
}e[N];
int f[110];

int getf(int x) {
    if (x != f[x]) f[x] = getf(f[x]);
    return f[x];
}

void merge(int x, int y) {
    int tx = getf(x);
    int ty = getf(y);
    if (tx != ty) {
        f[ty] = tx;
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    int n;
    while (cin >> n && n) {
        for (int i = 1; i <= n; i++) f[i] = i;
        int num = n * (n - 1) / 2;
        int cnt = 0, dis = 0;
        for (int i = 0; i < num; i++) {
            cin >> e[i].x >> e[i].y >> e[i].wei >> e[i].sta;
            if (e[i].sta == 1) merge(e[i].x, e[i].y);
        }
        for (int i = 1; i <= n; i++) {
            if (f[i] == i)
                cnt++;
        }
        if (cnt == 1) {
            cout << 0 << endl;
        } else {
            sort(e, e + num);
            for (int i = 0; i < num; i++) {
                if (getf(e[i].x) == getf(e[i].y)) continue;
                merge(e[i].x, e[i].y);
                dis += e[i].wei;
                if (--cnt == 1) {
                    cout << dis << endl;
                    break;
                }
            }
        }
    }

    return 0;
}

标签：cnt,const,上机,ty,--,wei,浙大,int,getf
来源： https://blog.csdn.net/weixin_43438720/article/details/123581732