（beginer） 最小生成树 UVA 10600 ACM Contest and Blackout

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

题意：找出最小生成树和次小生成树的长度。可能是相等的。

思路：首先找出最小生成树，然后在这个树中求出任意两点之间的最大边权，然后在枚举没有加进最小生成树的边中加边进来，假设加进来的边是(u,v) 那么我们从原来的的生成树中删掉u->v的路径中最大的边。

代码：

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 110;
const int inf = 1e9;
int n , m;
int p[maxn];
int maxcost[maxn][maxn];
bool vis[maxn];
bool sel[maxn*maxn];

int find(int x)
{
	if (x==p[x]) return x;
	return p[x] = find(p[x]);
}

struct Edge
{
	Edge(int uu=0,int vv=0,int ww=0) : u(uu) , v(vv) , w(ww) { }
	int u , v , w;
}edge[maxn*maxn];

inline bool operator < (const Edge &e1,const Edge &e2)
{
	return e1.w < e2.w;
}

vector<Edge> G[maxn];

inline int max(int a,int b) { return a < b ? b : a; }
void input()
{
	int u ,v , w;
	for (int i = 0 ; i < m ; ++i)
	{
		scanf("%d%d%d",&u,&v,&w);
		edge[i] = Edge(u,v,w);
	}
	sort(edge,edge+m);
}

void dfs(int s,int x,int maxc)
{
	if (vis[x]) return;
	vis[x] = true;
	if (maxcost[s][x] < maxc) maxcost[s][x] = maxc;
	for (int i = 0 ; i < G[x].size() ; ++i)
	{
		int y = G[x][i].v;
		dfs(s,y,max(maxc,G[x][i].w));
	}
}

void solve()
{
	for (int i = 1 ; i <= n ; ++i) p[i] = i , G[i].clear();
	int u , v , w;
	int S1 = 0 , S2 = inf;
	for (int i = 0 ; i < m ; ++i)
	{
		sel[i] = false;
		u = find(edge[i].u);
		v = find(edge[i].v);
		if (u==v) continue;
		sel[i] = true;
		p[u] = v;
		u = edge[i].u , v = edge[i].v;
		w = edge[i].w;
		G[u].push_back(Edge(u,v,w));
		G[v].push_back(Edge(v,u,w));
		S1 += w;
	}
	memset(maxcost,-1,sizeof(maxcost));
	printf("%d ",S1);
	for (int i = 1 ; i <= n ; ++i) { memset(vis,false,sizeof(vis)); dfs(i,i,-1); }
	for (int i = 0 ; i < m ; ++i) if (!sel[i])
		S2 = min(S2,S1+edge[i].w-maxcost[edge[i].u][edge[i].v]);
	printf("%d\n",S2);
}

int main()
{
	int T; cin>>T;
	while (T--)
	{
		scanf("%d%d",&n,&m);
		input();
		solve();
	}
}