HDU 6041.I Curse Myself 无向仙人掌图

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 V(k) as the weight of the k-th smallest weighted spanning tree of this graph, however, V(k) would be defined as zero if there did not exist k different weighted spanning trees.

Please calculate (∑k=1Kk⋅V(k))mod232.

 

Input

The input contains multiple test cases.

For each test case, the first line contains two positive integers n,m (2≤n≤1000,n−1≤m≤2n−3), the number of nodes and the number of edges of this graph.

Each of the next m lines contains three positive integers x,y,z (1≤x,y≤n,1≤z≤106), meaning an edge weighted z between node x and node y. There does not exist multi-edge or self-loop in this graph.

The last line contains a positive integer K (1≤K≤105).

 

Output

For each test case, output "Case #x: y" in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.

 

Sample Input

4 3

1 2 1

1 3 2

1 4 3

1

3 3

1 2 1

2 3 2

3 1 3

4

6 7

1 2 4

1 3 2

3 5 7

1 5 3

2 4 1

2 6 2

6 4 5

7

 

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
",++Case,sum);
    }
    return 0;
}