UVa 10600

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ACM Contest and Blackout

Solution Sketch

Suppose we have a set $S$ which represents all edges that are in MST, and a set $P$ represents all edges that are not in MST.

The crucial observation for getting the next-to-minimal spanning tree is: it can be obtained by adding one edge from $P$ and removing the edge from $S$ which caused the loop after adding an edge from $P$.

AC code

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#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> ii;
#define N 111
struct UFDS {
    int par[N];
    void init()
    {
        memset(par, -1, sizeof(par));
    }
    int root(int x)
    {
        return par[x] < 0 ? x : par[x] = root(par[x]);
    }
    void merge(int x, int y)
    {
        x = root(x);
        y = root(y);
        if (x == y)
            return;
        if (par[x] > par[y])
            swap(x, y);
        par[x] += par[y];
        par[y] = x;
    }
} ufds;
int main()
{
    int ncase;
    scanf("%d", &ncase);
    while (ncase--) {
        int n, m;
        scanf("%d %d", &n, &m);
        vector<pair<int, ii>> edge;
        for (int i = 0; i < m; i++) {
            int u, v, w;
            scanf("%d %d %d", &u, &v, &w);
            edge.push_back(make_pair(w, ii(u, v)));
        }
        sort(edge.begin(), edge.end());
        // get MST
        ufds.init();
        vector<int> mst, nmst; // used edge
        int ans1 = 0;
        for (int i = 0; i < (int)edge.size(); i++) {
            int u = edge[i].second.first;
            int v = edge[i].second.second;
            int w = edge[i].first;
            if (ufds.root(u) == ufds.root(v)) {
                nmst.push_back(i);
                continue;
            }
            ufds.merge(u, v);
            mst.push_back(i);
            ans1 += w;
            // printf("%d %d %d\n", i, w, ans1);
        }
        int ans2 = INT_MAX;
        if (n - 1 == m)
            ans2 = ans1;
        else {
            // change one selected edge at a time, and try to replace it with all
            // unselected edge
            for (int i = 0; i < n - 1; i++) {
                int tmp = 0;
                // redo ufds
                ufds.init();
                for (auto j : mst) {
                    if (mst[i] == j)
                        continue;
                    int u = edge[j].second.first;
                    int v = edge[j].second.second;
                    int w = edge[j].first;
                    ufds.merge(u, v);
                    tmp += w;
                }
                // try all unused edge
                for (auto j : nmst) {
                    int u = edge[j].second.first;
                    int v = edge[j].second.second;
                    int w = edge[j].first;
                    if (ufds.root(u) != ufds.root(v)) {
                        ans2 = min(ans2, tmp + w);
                    }
                }
            }
        }
        printf("%d %d\n", ans1, ans2);
    }
    return 0;
}