HDU 2121 Ice_cream’s world II (最小树形图+虚点，4级)

最新推荐文章于 2020-02-27 22:54:55 发布

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最新推荐文章于 2020-02-27 22:54:55 发布

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分类专栏：最小树形图

本文链接：https://blog.youkuaiyun.com/nealgavin/article/details/10772939

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在冰激凌世界中，女王授予土地给ACM者，并要求选择一个合适的地点作为首都。这个世界有N个城市和M条道路，每条道路都是有向的。Wiskey作为首席工程师，需要找到一个位置建立首都，使得从该位置出发可以访问每个城市，同时确保项目成本尽可能低。如果无法满足女王的要求，则输出“impossible”。反之，则输出最小成本和合适的城市编号。

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C - Ice_cream’s world II

Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2121

Description

After awarded lands to ACMers, the queen want to choose a city be her capital. This is an important event in ice_cream world, and it also a very difficult problem, because the world have N cities and M roads, every road was directed. Wiskey is a chief engineer in ice_cream world. The queen asked Wiskey must find a suitable location to establish the capital, beautify the roads which let capital can visit each city and the project’s cost as less as better. If Wiskey can’t fulfill the queen’s require, he will be punishing.

Input

Every case have two integers N and M (N<=1000, M<=10000), the cities numbered 0…N-1, following M lines, each line contain three integers S, T and C, meaning from S to T have a road will cost C.

Output

If no location satisfy the queen’s require, you must be output “impossible”, otherwise, print the minimum cost in this project and suitable city’s number. May be exist many suitable cities, choose the minimum number city. After every case print one blank.

Sample Input

Sample Output

     
      impossible

40 0
     
     

     
     
      思路：建一个虚节点，向其他所有点连边值比其他所有边和大，为sum，最后一定会得到最小树形图，判断这个最小树形图的值比sum+sum大则说明选了两条这种边，无答案
     
     
          否则答案就是其减去sum，边就是选了root 的边，由此可知道树根
     
     

     
     
      #include<cstdio>
#include<cstring>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
typedef long long type;
using namespace std;
const int nn=1005;
const int mm=17009;
const long long oo=1e18;
type sum;
class Edge
{
  public:int u,v;type w;
}e[mm];
int id[nn],vis[nn],pre[mm],city;
type in[nn];
type Directed_MST(int root,int V,int E)
{
  type ret=0;
  while(1)
  {
    FOR(i,0,V-1)in[i]=oo;
    in[root]=0;
    int u,v;
    FOR(i,0,E-1)
    {
      u=e[i].u;v=e[i].v;
      if(u!=v&&e[i].w<in[v])
      {
        in[v]=e[i].w;pre[v]=u;
        if(u==root)
          city=i;
      }
    }
    FOR(i,0,V-1)
    if(in[i]==oo)return -1;
    //find cicle
    int bcc_no=0;
    clr(vis,-1);clr(id,-1);
    FOR(i,0,V-1)
    { ret+=in[i];//if(in[i]==sum)city=i;
      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++;
    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 n,m;
  while(~scanf("%d%d",&n,&m))
  { sum=0;
    FOR(i,0,m-1)
    {
      scanf("%d%d%I64d",&e[i].u,&e[i].v,&e[i].w);
      sum+=e[i].w;
    }
    ++sum;
    FOR(i,0,n-1)
    e[i+m].u=n,e[i+m].v=i,e[i+m].w=sum;
    type ans=Directed_MST(n,n+1,n+m);
    //cout<<ans<<endl;
    ans-=sum;
    if(ans>=sum)printf("impossible\n");
    else printf("%I64d %d\n",ans,city-m);
    printf("\n");
  }
  return 0;
}