UVA 11183 Teen Girl Squad （最小树形图，3级）

最新推荐文章于 2019-01-08 20:35:27 发布

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本文深入探讨了游戏开发领域的关键技术，包括游戏引擎、编程语言、硬件优化等，并重点介绍了AI音视频处理在游戏场景中的应用，如语义识别、语音识别、图像处理等，为游戏开发者提供了一站式的解决方案和技术指南。

B - Teen Girl Squad

Crawling in process... Crawling failed Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

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Appoint description:

Description

Problem I
Teen Girl Squad
Input: Standard Input

Output: Standard Output

-- 3 spring rolls please.
-- MSG'D!!
-- Oh! My stomach lining!

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

0 1 10

1 0 10

0 1 10

0 2 10

1 3 20

2 3 30

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;
}