uva 10600 - ACM Contest and Blackout（次小生成树）

In order to prepare the “The First National ACM School Contest”(in 20??) the major of the city decided to provide all the schools with a reliable source of power. (The major is really afraid of blackoutsJ). So, in order to do that, power station “Future” and one school (doesn’t matter which one) must be connected; in addition, some schools must be connected as well.

You may assume that a school has a reliable source of power if it’s connected directly to “Future”, or to any other school that has a reliable source of power. You are given the cost of connection between some schools. The major has decided to pick out two the cheapest connection plans – the cost of the connection is equal to the sum of the connections between the schools. Your task is to help the major – find the cost of the two cheapest connection plans.

Input

The Input starts with the number of test cases, T (1£T£15) on a line. Then T test cases follow. The first line of every test case contains two numbers, which are separated by a space, N (3£N£100) the number of schools in the city, and M the number of possible connections among them. Next M lines contain three numbers A_i, B_i, C_i , where C_i is the cost of the connection (1£C_i£300) between schools A_i and B_i. The schools are numbered with integers in the range 1 to N.

Output

For every test case print only one line of output. This line should contain two numbers separated by a single space - the cost of two the cheapest connection plans. Let S₁ be the cheapest cost and S₂ the next cheapest cost. It’s important, that S₁=S₂ if and only if there are two cheapest plans, otherwise S₁£S₂. You can assume that it is always possible to find the costs S₁ and S₂..

Sample Input

Sample Output

2

5 8

1 3 75

3 4 51

2 4 19

3 2 95

2 5 42

5 4 31

1 2 9

3 5 66

9 14

1 2 4

1 8 8

2 8 11

3 2 8

8 9 7

8 7 1

7 9 6

9 3 2

3 4 7

3 6 4

7 6 2

4 6 14

4 5 9

5 6 10

110 121

37 37

Problem source: Ukrainian National Olympiad in Informatics 2001

Problem author: Shamil Yagiyayev

Problem submitter: Dmytro Chernysh

Problem solution: Shamil Yagiyayev, Dmytro Chernysh, K M Hasan

裸的次小生成树。求法，用vised记录在最小生成树中的边，然后枚举所有没有被vised过得，进行替换，找长度最小的替换。

#include<cstdio>#include<cmath>#include<cstring>#include<vector>#include<algorithm>using namespace std;const int maxn = 100 + 10;const int INF = 1000000000;struct Edge {  int x, y;  double d;  bool operator < (const Edge& rhs) const {    return d < rhs.d;  }};struct MST{    int n, m;//点数和边数    Edge e[maxn*maxn];//储存所有的边    int pa[maxn];//用于并查集，父指针    vector<int> G[maxn];//用于Dfs，储存生成树中每个点相邻的点    vector<int> C[maxn];//用于Dfs，储存相应的边的权    vector<int> nodes;//用于Dfs，储存已经遍历过的节点    int maxcost[maxn][maxn];//储存最小生成树中，u、v唯一路径上的最大权值    int used[maxn*maxn];    //这里是把无根树，转为了有根树，具体做法：    //在初始调用时Dfs(0,-1,0)，然后后面节点拓展儿子时，不允许向回走    void dfs(int u, int fa, int facost){        //先对当前点同所有已访问过的节点，进行更新        for(int i = 0; i < nodes.size(); i++) {            int x = nodes[i];            maxcost[u][x] = maxcost[x][u] = max(maxcost[x][fa], facost);        }        nodes.push_back(u);        //再递归下一个点        for(int i = 0; i < G[u].size(); i++) {            int v = G[u][i];            if(v != fa) dfs(v, u, C[u][i]);        }    }    void init(int n){        this -> n = n;        m = 0;        memset(maxcost,0,sizeof(maxcost));        memset(used,0,sizeof(used));        nodes.clear();        for(int i = 0; i < n; i++) { pa[i] = i; G[i].clear(); C[i].clear(); }    }    void AddEdge(int x,int y,int dist){        e[m++] = (Edge){x,y,dist};    }    int findset(int x) { return pa[x] != x ? pa[x] = findset(pa[x]) : x; }    int solve() {        sort(e, e+m);        int cnt = 0;        int ans = 0;        for(int i = 0; i < m; i++) {            int x = e[i].x, y = e[i].y, u = findset(x), v = findset(y);            int d = e[i].d;                if(u != v) {                    used[i] = 1;                    pa[u] = v;                    G[x].push_back(y); C[x].push_back(d);                    G[y].push_back(x); C[y].push_back(d);                    ans += d;                    if(++cnt == n-1) break;                }            }        return ans;    }};MST solver;int main(){    int t,n,m;    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&m);        solver.init(n);        for(int i = 0;i < m;i++){            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            solver.AddEdge(a-1,b-1,c);        }        int ans1 = solver.solve();        int ans2 = INF;        solver.dfs(0,-1,0);        for(int i = 0;i < m;i++){            if(!solver.used[i]){                int x = solver.e[i].x;                int y = solver.e[i].y;                int d = solver.e[i].d;                ans2 = min(ans2,ans1+d-solver.maxcost[x][y]);            }        }        printf("%d %d\n",ans1,ans2);    }    return 0;}