uva10600 - ACM Contest and Blackout 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 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 |
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 |
求最小生成树和次小生成树。
次小生成树：先求最小生成树，记录最小生成树上的边，然后用DFS（BFS也行）求出任意两点u,v在最小生成树上唯一路径上的最大边maxe[u][v]，那么只要枚举增加一条不在最小生成树上的边u,v，并且删除u-v在最小生成树路径上的最大边maxe[u][v]（增加边后形成环，删掉一条边后任意两点仍然相互可达），在枚举的里面找到一个最小的生成树就是次小生成树。很明显次小生成树是最小生成树删掉一条边加上一条新边，不可能删掉两条，没有删掉一条划得来。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<vector>
using namespace std;
typedef pair<int,int> pii;
const int MAXN=110;
const int MAXM=5010;
const int MAXNODE=4*MAXN;
const int LOGMAXN=50;
const int INF=0x3f3f3f3f;
int T,N,M;
int mst[MAXN][MAXN],maxe[MAXN][MAXN],pa[MAXN];
vector<int> G[MAXN],C[MAXN],node;
int find(int x){
return x==pa[x]?x:pa[x]=find(pa[x]);
}
struct Edge{
int u,v,d;
bool operator < (const Edge& rhs) const{
return d<rhs.d;
}
}e[MAXM];
int MST(){
sort(e,e+M);
memset(mst,0,sizeof(mst));
for(int i=0;i<=N;i++){
pa[i]=i;
G[i].clear();
C[i].clear();
}
int cnt=0,ret=0;
for(int i=0;i<M;i++){
int x=find(e[i].u),y=find(e[i].v);
if(x!=y){
pa[x]=y;
G[e[i].u].push_back(e[i].v);
C[e[i].u].push_back(e[i].d);
G[e[i].v].push_back(e[i].u);
C[e[i].v].push_back(e[i].d);
mst[e[i].u][e[i].v]=mst[e[i].v][e[i].u]=1;
cnt++;
ret+=e[i].d;
}
if(cnt==N-1) return ret;
}
return -1;
}
void DFS(int u,int fa,int d){
int len=node.size();
for(int i=0;i<len;i++){
int v=node[i];
maxe[u][v]=maxe[v][u]=max(maxe[v][fa],d);
}
len=G[u].size();
node.push_back(u);
for(int i=0;i<len;i++){
int v=G[u][i];
if(v!=fa) DFS(v,u,C[u][i]);
}
}
int main(){
freopen("in.txt","r",stdin);
scanf("%d",&T);
while(T--){
scanf("%d%d",&N,&M);
int u,v,d;
for(int i=0;i<M;i++) scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].d);
int ans1=MST();
node.clear();
memset(maxe,-1,sizeof(maxe));
DFS(1,-1,0);
int ans2=INF;
for(int i=0;i<M;i++) if(!mst[e[i].u][e[i].v]){
ans2=min(ans2,ans1+e[i].d-maxe[e[i].u][e[i].v]);
}
printf("%d %d\n",ans1,ans2);
}
return 0;
}