UVA-10600.Contest and Blackout.(Kruskal + 次小生成树)

题目链接

本题思路：模版的次小生成树问题，输出MST and Second_MST的值。

参考代码：

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

const int maxn = 100 + 5, maxe = 100 * 100 / 2 + 5, INF = 0x3f3f3f3f;
int n, m, Max[maxn][maxn], pir[maxn];
struct Edge {
    int u, v, w;
    bool vis;
}edge[maxe];
vector<int> G[maxn];

bool cmp(const Edge &a, const Edge &b) {
    return a.w < b.w;
}

int Find(int x) {
    if(x == pir[x]) return x;
    return pir[x] = Find(pir[x]);
}

int Kruskal() {
    sort(edge + 1, edge + m + 1, cmp);
    for(int i = 1; i <= n; i ++) {
        G[i].clear();
        pir[i] = i;
        G[i].push_back(i);
    }
    int cnt = 0, ans = 0;
    for(int i = 1; i <= m; i ++) {
        int fx = Find(edge[i].u), fy = Find(edge[i].v);
        if(cnt == n - 1) break;
        if(fx != fy) {
            cnt ++;
            int len_fx = G[fx].size(), len_fy = G[fy].size();
            edge[i].vis = true;
            ans += edge[i].w;
            for(int j = 0; j < len_fx; j ++) {
                for(int k = 0; k < len_fy; k ++) {
                    Max[G[fx][j]][G[fy][k]] = Max[G[fy][k]][G[fx][j]] = edge[i].w;
                }
            }
            pir[fx] = fy;
            for(int j = 0; j < len_fx; j ++)
                G[fy].push_back(G[fx][j]);
        }
    }
    if(cnt < n - 1) return INF;
    else return ans;
}

int Second_Kruskal(int MST) {
    int ans = INF;
    for(int i = 1; i <= m; i ++) {
        if(!edge[i].vis)
            ans = min(ans, MST + edge[i].w - Max[edge[i].u][edge[i].v]);
    }
    return ans;
}

int main () {
    int t;
    scanf("%d", &t);
    while(t --) {
        scanf("%d %d", &n, &m);
        for(int i = 1; i <= m; i ++) {
            scanf("%d %d %d", &edge[i].u, &edge[i].v, &edge[i].w);
            edge[i].vis = false;
        }
        int MST = Kruskal();
        int Second_MST = Second_Kruskal(MST);
        printf("%d %d
", MST, Second_MST);
    }
    return 0;
}