AcWing 1140. 最短网络

题目传送门

一个很祼的最小生成树问题，出发点已确定:

#include 

using namespace std;
const int N = 510;
const int INF = 0x3f3f3f3f;

int n, m;
int g[N][N];
int dist[N];
bool st[N];
int res;

int prim() {
    memset(dist, 0x3f, sizeof dist);
    for (int i = 0; i < n; i++) {
        int t = -1;
        for (int j = 1; j <= n; j++)
            if (!st[j] && (t == -1 || dist[t] > dist[j])) t = j;
        if (i && dist[t] == INF) return INF;
        if (i) res += dist[t];
        for (int j = 1; j <= n; j++) dist[j] = min(dist[j], g[t][j]);
        st[t] = true;
    }
    return res;
}

int main() {
    cin >> n;
    //完全图，每两个点之间都有距离，不用考虑无解情况
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            cin >> g[i][j];

    int t = prim();
    printf("%d\n", t);
    return 0;
}