HDU 6041.I Curse Myself 无向仙人掌图

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原文链接：http://www.cnblogs.com/GeekZRF/p/7671257.html

文章标签：

#java #php

本文探讨了一个关于仙人掌图的生成树问题，即如何找到权重最小的k个生成树。通过分析仙人掌图的特点，文章提供了一种有效的解决方案，并给出了详细的代码实现。

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I Curse Myself

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2266 Accepted Submission(s): 544

Problem Description

There is a connected undirected graph with weights on its edges. It is guaranteed that each edge appears in at most one simple cycle.

Assuming that the weight of a weighted spanning tree is the sum of weights on its edges, define

Input

The input contains multiple test cases.

For each test case, the first line contains two positive integers

Output

For each test case, output " Case #" in one line (without quotes), where

Sample Input

4 3

1 2 1

1 3 2

1 4 3

3 3

1 2 1

2 3 2

3 1 3

6 7

1 2 4

1 3 2

3 5 7

1 5 3

2 4 1

2 6 2

6 4 5

Sample Output

Case #1: 6

Case #2:26

Case #3: 493

Source

2017 Multi-University Training Contest - Team 1

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题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=6041

题意：有一个n个结点，m条无向边的仙人掌图。求删除一些边形成生成树，求前k小生成树。

思路：因为是一个仙人掌图，所以每条边最多在一个简单环内。所以只需要删除每个简单环内的一条边就能形成生成树。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<bitset>
using namespace std;
#define PI acos(-1.0)
#define eps 1e-8
typedef long long ll;
typedef pair<int,int> P;
const int N=1e3+100,M=4e3+100;
struct edge
{
    int from,to;
    int w;
    int next;
};
int n,m,k;
edge es[M];
int cnt,head[N];
int dfs_clock=0;
int pre[N],low[N];
stack<int>s;
void init(int n)
{
    cnt=0;
    dfs_clock=0;
    for(int i=0; i<=n+10; i++) head[i]=-1,pre[i]=0;
}
void addedge(int u,int v,int w)
{
    cnt++;
    es[cnt].from=u,es[cnt].to=v;
    es[cnt].w=w;
    es[cnt].next=head[u];
    head[u]=cnt;
}
int tmp[100100],ans[100100];
struct node
{
    int num;
    int id;
    bool operator <(const node x) const
    {
        return x.num>num;
    }
};
void unit(priority_queue<node> &q)
{
    tmp[0]=0;
    while(tmp[0]<k&&!q.empty())
    {
        node x=q.top();
        q.pop();
        tmp[++tmp[0]]=x.num;
        if(++x.id<=ans[0]) q.push((node)
        {
            x.num-ans[x.id-1]+ans[x.id],x.id
        });
    }
    ans[0]=0;
    for(int i=1; i<=tmp[0]; i++) ans[++ans[0]]=tmp[i];
}
bool dfs(int u,int fa)
{
    pre[u]=low[u]=++dfs_clock;
    for(int i=head[u]; i!=-1; i=es[i].next)
    {
        int v=es[i].to;
        if(v==fa) continue;
        if(!pre[v])
        {
            s.push(i);
            dfs(v,u);
            low[u]=min(low[u],low[v]);
            if(low[v]>=pre[u])
            {
                priority_queue<node>q;
                while(!s.empty())
                {
                    int poi=s.top();
                    s.pop();
                    q.push((node){ans[1]+es[poi].w,1});
                    if(poi==i) break;
                }
                if(q.size()>1) unit(q);
            }
        }
        else if(pre[v]<pre[u]&&v!=fa)
        {
            s.push(i);
            low[u]=min(low[u],pre[v]);
        }
    }
}
int main()
{
    int Case=0;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init(n);
        int all=0;
        for(int i=1; i<=m; i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            all+=w;
            addedge(u,v,w),addedge(v,u,w);
        }
        scanf("%d",&k);
        ans[0]=0,ans[++ans[0]]=0;
        dfs(1,0);
        ll sum=0,mod=(1LL<<32);
        for(int i=1; i<=ans[0]; i++)
            sum=(sum+(1LL*(all-ans[i])*i)%mod)%mod;
        printf("Case #%d: %lld\n",++Case,sum);
    }
    return 0;
}