ACM Contest and Blackout
题目链接：https://vjudge.net/problem/UVA-10600
Description:
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 Ai , Bi , Ci , where Ci is the cost of the connection (1 < Ci < 300) between schools Ai and Bi . 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:
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
Sample Output:
110 121
37 37
题意：
求次小生成树。
题解：
先跑一遍最小生成树，然后O(n^2)预处理出任意两点之间的最小瓶颈路，最后通过枚举算出次小生成树。
代码如下：
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <queue> #include <cmath> #define INF 0x3f3f3f3f using namespace std; typedef long long ll; const int N = 105; int t,n,m; struct Edge{ int u,v,w; bool operator < (const Edge &A)const{ return w<A.w; } }e[N*N]; int f[N],mp[N][N]; int find(int x){ return f[x]==x?f[x]:f[x]=find(f[x]); } int Kruskal(){ int ans=0; for(int i=0;i<=n+1;i++) f[i]=i; for(int i=1;i<=m;i++){ int u=e[i].u,v=e[i].v; int fx=find(u),fy=find(v); if(fx==fy) continue ; f[fx]=fy; mp[u][v]=mp[v][u]=1; ans+=e[i].w; } return ans ; } int d[N][N],dis[N][N]; int check[N]; void dfs(int u,int fa){ for(int i=1;i<=n;i++){ if(check[i]) d[i][u]=d[u][i]=max(d[i][fa],dis[u][fa]); } check[u]=1; for(int i=1;i<=n;i++){ if(mp[i][u] && i!=fa) dfs(i,u); } } int main(){ scanf("%d",&t); while(t--){ scanf("%d%d",&n,&m); memset(dis,0,sizeof(dis)); for(int i=1;i<=m;i++){ int u,v,w; scanf("%d%d%d",&u,&v,&w); e[i]=Edge{u,v,w}; dis[u][v]=dis[v][u]=w; } sort(e+1,e+m+1); memset(d,0,sizeof(d)); memset(check,0,sizeof(check)); memset(mp,0,sizeof(mp)); int sum=Kruskal(); cout<<sum<<" "; dfs(1,-1); int ans=INF; for(int i=1;i<=m;i++){ int u=e[i].u,v=e[i].v,w=e[i].w; if(mp[u][v]) continue ; ans=min(ans,sum-d[u][v]+w); } cout<<ans<<endl; } return 0; }