HDU 6041 I Curse Myself （仙人掌图）

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 (∑Kk=1k⋅V(k))mod 232 .

 

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 31 2 11 3 21 4 313 31 2 12 3 23 1 346 71 2 41 3 23 5 71 5 32 4 12 6 26 4 57

 

Sample Output

Case #1: 6Case #2: 26Case #3: 493

 

题意

求一棵无向仙人掌图中前 k 小的生成树权值和。

 

思路

首先我们需要清楚什么是无向仙人掌图，它是一张连通图，且任意一条边都至多属于一个环。

也就是说，仙人掌图的一棵生成树就是其每个环去掉一条边所形成的图。

那么我们可以一次通过 dfs 找出所有环，然后存储每个环中的所有边长，此时问题就变成了在 n 个数组中各取一个数，求其和的前 k 大的数，也算一个经典问题啦~

PS：在最后求解前 k 大的和时千万不要用 O(n3) 的算法，否则 TLE 到 cry ��

 

AC 代码

#include<cstdio>#include<cstdlib>#include<cstring>#include<stdlib.h>#include<iostream>#include<queue>#include<vector>#include<map>#include<cmath>#include<algorithm>using namespace std;const int maxn = 2100;typedef pair<int,int> P;typedef __int64 LL;struct node{    int to;    int next;    int cost;} edge[maxn<<1];int head[maxn<<1],tot;int Stack[maxn],stop;bool instack[maxn];int n,m,k;bool isStart;bool vis[maxn];int g[maxn][maxn];int Bcnt;vector<int>G[maxn];vector<int> f,tmpf;LL mod = 1LL<<32;void init(){    memset(head,-1,sizeof(head));    memset(vis,false,sizeof(vis));    memset(instack,false,sizeof(instack));    tot=stop=Bcnt=0;    isStart=false;    f.clear();    tmpf.clear();}void addedge(int u,int v,int cost){    edge[tot].to=v;    edge[tot].next=head[u];    edge[tot].cost=cost;    head[u]=tot++;}void dfs(int x,int fa){    Stack[++stop]=x;    instack[x]=true;    vis[x]=true;    for(int i=head[x]; i!=-1; i=edge[i].next)    {        int to=edge[i].to;        if(to==fa)continue;        if(!vis[to])            dfs(to,x);        else if(instack[to])        // 找到仙人掌的一个环        {            vector<int> G;            int top=stop;            int p=to;            while(Stack[top]!=to)   // 统计环中所有边长            {                G.push_back(g[p][Stack[top]]);                p=Stack[top--];            }            G.push_back(g[p][Stack[top]]);            sort(G.begin(),G.end(),[](const int &a,const int &b)    // 边长从大到小排序            {                return a>b;            });            int len=G.size();            if(!isStart)            // 第一次计算            {                for(int j=0; j<len; j++)                    f.push_back(G[j]);                isStart=true;            }            else            {                int cnt=f.size()*len;                cnt=min(cnt,k);                priority_queue<P> que;                for(int j=0; j<len; j++)                    que.push(P(f[0]+G[j],0));                tmpf.clear();                while(cnt--)                {                    P p=que.top();                    que.pop();                    tmpf.push_back(p.first);                    if(p.second!=(int)f.size()-1)                    {                        p.first-=f[p.second++];                        p.first+=f[p.second];                        que.push(p);                    }                }                swap(f,tmpf);            }        }    }    instack[x]=false;    stop--;}int main(){    for(int ti=1; ~scanf("%d%d",&n,&m); ti++)    {        init();        LL cnt=0,ans=0;        for(int i=0; i<m; i++)        {            int u,v,cost;            cin>>u>>v>>cost;            addedge(u,v,cost);            addedge(v,u,cost);            g[u][v]=g[v][u]=cost;            cnt+=cost;        }        scanf("%d",&k);        dfs(1,-1);        int len=f.size();        if(!len)ans=cnt;        for(int i=0; i<len; i++)            ans=(ans+(1LL*(i+1)*(cnt-f[i]))%mod)%mod;        printf("Case #%d: %I64d\n",ti,ans);    }    return 0;}