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  • 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=1KkV(k))mod232.
     
    Input
    The input contains multiple test cases.

    For each test case, the first line contains two positive integers n,m (2n1000,n1m2n3), 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 (1x,yn,1z106), 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 (1K105).
     
    Output
    For each test case, output "Case #xy" 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
     
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    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;
    }
    无向仙人掌图
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  • 原文地址:https://www.cnblogs.com/GeekZRF/p/7671257.html
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