UVA - 10600 ACM Contest and Blackout(次小生成树)-优快云博客

本文介绍了一种求解最小生成树及其次小生成树的方法。首先通过Prim算法找到最小生成树，然后通过特殊技巧确定次小生成树的权值。文章详细展示了整个过程的代码实现。

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题目大意：给出一张图，问这张图的最小生成树的权值和次小生成树的权值，如果有两个最小生成树，则输出的权值都为最小生成树的权值

解题思路：先找出最小生成树，再标记一下形成最小生成树的树边，接着枚举那些不是树边的边

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

#define N 110
#define M 10010
#define INF 0x3f3f3f3f

struct Node{
    int x, y;
}node[M];

int dis[N][N], d[N], f[N], maxcost[N][N];
int n, m;
bool mark[N], vis[N][N];
vector<int> v;

void init() {
    scanf("%d%d", &n, &m);
    memset(dis, 0x3f, sizeof(dis));
    for (int i = 1; i <= n; i++)
        dis[i][i] = 0;

    int u, v, c;
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &u, &v, &c);
        node[i].x = u; node[i].y = v;;
        dis[u][v] = dis[v][u] = min(dis[u][v], c);
    }
}

int Prim() {
    for (int i = 1; i <= n; i++)
        d[i] = INF;
    memset(mark, 0, sizeof(mark));
    memset(vis, 0, sizeof(vis));
    d[1] = 0;
    f[1] = 1;
    v.clear();
    int ans = 0;
    for (int i = 1; i <= n; i++) {
        int t = INF, x;
        for (int j = 1; j <= n; j++) 
            if (!mark[j] && d[j] < t)
                t = d[x = j];

        mark[x] = vis[x][f[x]] = vis[f[x]][x] = true;
        ans += dis[f[x]][x];

        int size = v.size();
        for (int j = 0; j < size; j++) {
            maxcost[v[j]][x] = maxcost[x][v[j]] = max(maxcost[v[j]][f[x]], dis[f[x]][x]);
        }
        v.push_back(x);

        for (int j = 1; j <= n; j++) {
            if (!mark[j] && dis[x][j] < d[j]) {
                d[j] = dis[x][j];
                f[j] = x;
            }
        }
    }
    return ans;
}

void solve() {
    int ans = Prim();
    int ans2 = INF;

    int x, y;
    for (int i = 0; i < m; i++) {
        x = node[i].x;
        y = node[i].y;
        if (!vis[x][y]) {
            ans2 = min(ans2, ans - maxcost[x][y] + dis[x][y]);
        }
    }
    printf("%d %d\n", ans, ans2);
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}