URAL 1416 Confidential（次小生成树）

题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1416

Zaphod Beeblebrox — President of the Imperial Galactic Government. And by chance he is an owner of enterprises that trade in secondhand pens. This is a complicated highly protable and highly competitive business. If you want to stay a leader you are to minimize your expenses all the time. And the presedent's high post helps in those aairs. But he is to keep this business in secret. As a president Zaphod has access to the top secret and important information an exact value of power loss in the hyperspace transition between the planets. Of course, this information is very useful to his company. Zaphod is to choose the minimal possible set of trans-planet passages so that he could pass from any planet to any other one via those passages and their total cost would be minimal. The task won't be complicated if Zaphod was not to keep in secret that he helps his company with the secret information. Thus, Zaphod decided to find not the cheapest passages set but the next one. As a real businessman he wants to estimate the value of his conspiracy expenses.

Input

The first input line contains two integers: N (2 ≤ N ≤ 500) is a number of planets in the Galaxy and M is an amount of possible transitions. The next M lines contain three integers ai, bi the numbers of the planets that are connected with some passage (1 ≤ ai, bi ≤ N), and wi (0 ≤ wi ≤ 1000) is the transition cost. If an A to B transition is possible then a B to A transition is possible, too. The cost of those transitions are equal. There is not more than one passage between any two planets. One can reach any planet from any other planet via some chain of these passages.

Output

You should find two different sets of transitions with the minimal possible cost and output theirs costs. Print the minimal possible cost first. If any of those sets of transitions does not exist denote it's cost by −1.

题目大意：给一个n个点m条边的无向图，求最小生成树和次小生成树（图无重边，似乎是连通图）。

思路：参考2014年汪汀的IOI集训队论文《最小生成树问题的拓展》。时间复杂度为O(n^2)。

代码（0.109MS）：

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

 const int MAXV = ;
 const int MAXE = MAXV * MAXV;
 const int INF = 0x3f3f3f3f;

 int head[MAXV], ecnt;
 int to[MAXE], next[MAXE], cost[MAXE];
 bool select[MAXE];
 int n, m;

 void init() {
     memset(head + , -, n * sizeof(int));
     ecnt = ;
 }

 void add_edge(int u, int v, int c) {
     to[ecnt] = v; cost[ecnt] = c; next[ecnt] = head[u]; head[u] = ecnt++;
     to[ecnt] = u; cost[ecnt] = c; next[ecnt] = head[v]; head[v] = ecnt++;
 }

 int dis[MAXV], pre[MAXV];
 int maxPath[MAXV][MAXV];
 bool vis[MAXV];

 int prim() {
     memset(dis + , 0x3f, n * sizeof(int));
     dis[] = ;
     int res = ;
     for(int _ = ; _ < n; ++_) {
         int u = -;
         for(int i = ; i <= n; ++i) if(!vis[i] && dis[i] < INF)
             if(u == - || dis[i] < dis[u]) u = i;
         if(u == -) return -;
         for(int p = head[u]; ~p; p = next[p]) {
             int &v = to[p];
             if(vis[v]) continue;
             if(cost[p] < dis[v]) dis[v] = cost[p], pre[v] = p;
         }
         for(int i = ; i <= n; ++i) if(vis[i]) {
             int &v = to[pre[u] ^ ];
             maxPath[i][u] = maxPath[u][i] = max(maxPath[i][v], dis[u]);
         }
         res += dis[u];
         vis[u] = true;
         if(u != ) select[pre[u]] = select[pre[u] ^ ] = true;
     }
     return res;
 }

 int solve(int ans) {
     int res = -;
     for(int u = ; u <= n; ++u) {
         for(int p = head[u]; ~p; p = next[p]) {
             int &v = to[p];
             if(select[p] || u == v) continue;
             if(res == - || ans - maxPath[u][v] + cost[p] < res)
                 res = ans - maxPath[u][v] + cost[p];
         }
     }
     return res;
 }

 int main() {
     scanf("%d%d", &n, &m);
     init();
     for(int i = , a, b, c; i < m; ++i) {
         scanf("%d%d%d", &a, &b, &c);
         add_edge(a, b, c);
     }
     int ans1 = prim(), ans2 = -;
     if(ans1 != -) ans2 = solve(ans1);
     printf("Cost: %d\n", ans1);
     printf("Cost: %d\n", ans2);
 }