UVA 11183 Teen Girl Squad （最小树形图，3级）
Description
Problem I
Teen Girl Squad
Input: Standard Input
Output: Standard Output
| -- 3 spring rolls please. |
Strong Bad
You are part of a group of n teenage girls armed with cellphones. You have some news you want to tell everyone in the group. The problem is that no two of you are in the same room, and you must communicate using only cellphones. What's worse is that due to excessive usage, your parents have refused to pay your cellphone bills, so you must distribute the news by calling each other in the cheapest possible way. You will call several of your friends, they will call some of their friends, and so on until everyone in the group hears the news.
Each of you is using a different phone service provider, and you know the price of girl A calling girl B for all possible A and B. Not all of your friends like each other, and some of them will never call people they don't like. Your job is to find the cheapest possible sequence of calls so that the news spreads from you to all n-1 other members of the group.
Input
The first line of input gives the number of cases, N (N<150). N test cases follow. Each one starts with two lines containing n (0<= n<=1000) and m (0< = m <= 40,000) . Girls are numbered from 0 to n-1 , and you are girl 0. The next m lines will each contain 3 integers, u, v and w, meaning that a call from girl u to girl v costs w cents (0 <= w <= 1000) . No other calls are possible because of grudges, rivalries and because they are, like, lame. The input file size is around 1200 KB.
Output
For each test case, output one line containing "Case #x:" followed by the cost of the cheapest method of distributing the news. If there is no solution, print "Possums!" instead.
Sample Input Output for Sample Input
4 2 1 0 1 10 2 1 1 0 10 4 4 0 1 10 0 2 10 1 3 20 2 3 30 4 4 0 1 10 1 2 20 2 0 30 2 3 100 |
Case #1: 10 Case #2: Possums! Case #3: 40 Case #4: 130 |
Problem setter: Igor Naverniouk
思路：就是个模板最小树形图 80MS+
#include<iostream>
#include<cstring>
#include<cstdio>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int nn=1003;
const int mm=40005;
const int oo=1e5;
class Edge
{
public:int u,v,w;
}e[mm];
int in[nn],vis[nn],id[nn];
int pre[mm];
int Direct_MST(int root,int V,int E)
{
int ret=0;
while(1)
{
FOR(i,0,V-1)in[i]=oo;
in[root]=0;
FOR(i,0,E-1)
{
int u=e[i].u,v=e[i].v;
while(u!=v&&e[i].w<in[v])
{
in[v]=e[i].w;pre[v]=u;
}
}//min in edge
FOR(i,0,V-1)
{
if(i==root)continue;
if(in[i]==oo)return -1;
}
clr(vis,-1);clr(id,-1);
//find cicle
int bcc_no=0;
FOR(i,0,V-1)
{
ret+=in[i];
int v=i;
while(vis[v]!=i&&id[v]==-1&&v!=root)
vis[v]=i,v=pre[v];
if(v!=root&&id[v]==-1)
{for(int u=pre[v];u!=v;u=pre[u])
id[u]=bcc_no;
id[v]=bcc_no++;
}
}
if(bcc_no==0)break;
FOR(i,0,V-1)
if(id[i]==-1)id[i]=bcc_no++;
//缩点后处理
int u,v;
FOR(i,0,E-1)
{
u=e[i].u;v=e[i].v;
e[i].u=id[u];
e[i].v=id[v];
if(id[u]^id[v])
e[i].w-=in[v];
}
V=bcc_no;
root=id[root];
}
return ret;
}
int main()
{
int cas,n,m;
while(~scanf("%d",&cas))
{
FOR(ca,1,cas)
{
scanf("%d%d",&n,&m);
FOR(i,0,m-1)
{
scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
}
int ans=Direct_MST(0,n,m);
if(ans==-1)printf("Case #%d: Possums!\n",ca);
else printf("Case #%d: %d\n",ca,ans);
}
}
return 0;
}