uva 10600 次小生成树

Problem A

ACM contest and Blackout

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

 

求两棵最小的生成树权值，第一个显然mst，第二个用n^2 dfs求出任意两点间最大边权，然后枚举没有在mst中的边求次小mst即可

1A代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<map>

using namespace std;

#define LL long long
#define ULL unsigned long long
#define UINT unsigned int
#define MAX_INT 0x7fffffff
#define cint const int
#define MAX(X,Y) ((X) > (Y) ? (X) : (Y))
#define MIN(X,Y) ((X) < (Y) ? (X) : (Y))

#define MAXN 111
#define MAXM 11111
#define INF 100000000

struct edge{
    int u, v, d;
    bool operator < (const edge &rhs)const{
        return d < rhs.d;
    }
}e[MAXM];
vector<int> g[MAXN];
int n, m;

int p[MAXN];
int finds(int x){
    if(p[x]==-1) return x;
    else return (p[x]=finds(p[x]));
}

bool vis[MAXM];
int w[MAXN][MAXN];
int mst(){
    fill_n(p+1, n, -1);
    fill_n(vis, m, false);
    for(int i=1; i<=n; i++) g[i].clear();
    sort(e, e+m);

    int ans=0;
    for(int i=0, j=0; i<m; i++){
        int x = e[i].u, y = e[i].v, d = e[i].d;
        int fx = finds(x), fy = finds(y);
        if(fx!=fy){
            p[fx]=fy;
            ans+=d;                 vis[i]=true;
            g[x].push_back(y);      w[x][y]=w[y][x]=d;
            g[y].push_back(x);
            if(++j==n-1) break;
        }
    }
    return ans;
}

int maxcst[MAXN][MAXN];
vector<int> pre;
//bool viv[MAXN];

void dfs(int u, int fa){
//    viv[u]=true;
    for(int i=0; i<pre.size(); i++){
        int v = pre[i], d = w[fa][u];
        maxcst[u][v]=maxcst[v][u]=
        MAX(d, maxcst[fa][v]);
    }
    pre.push_back(u);
    for(int i=0; i<g[u].size(); i++) if(fa!=g[u][i])
        dfs(g[u][i], u);
}

int main(){
//    freopen("C:\Users\Administrator\Desktop\in.txt","r",stdin);
    int T;
    scanf(" %d", &T);
    while(T--){
        int i;
        scanf(" %d %d", &n, &m);
        for(i=0; i<m; i++){
            int u, v, w;
            scanf(" %d %d %d", &u, &v, &w);
            if(u!=v) e[i]=(edge){u, v, w};       // - -
        }

        int fir = mst();
        for(i=1; i<=n; i++) fill_n(maxcst[i]+1, n, -1);
//        fill_n(viv+1, n, false);
        pre.clear();
//        printf("%d
", fir);
        dfs(1, -1);
        int sec = INF;
        for(i=0; i<m; i++) if(!vis[i]){
            int u = e[i].u, v = e[i].v, d = e[i].d;
            sec=MIN(sec, fir-maxcst[u][v]+d);
//            cout<<"^^^^^^^^^^^^^^^^^^^^^^"<<endl;
//            cout<<(fir-maxcst[u][v]+d)<<endl;
//            cout<<maxcst[u][v]<<endl;
////            cout<<e[i].u<<' '<<e[i].v<<' '<<e[i].d<<endl;
//            cout<<"&&&&&&&&&&&&&&&&&&&&&&&"<<endl;
        }
        printf("%d %d
", fir, sec);
    }
    return 0;
}