hdu 6041 I Curse Myself 无向图找环+优先队列

I Curse Myself

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)



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

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
#define PI acosI(-1.0)
typedef long long ll;
typedef pair<int,int> P;
const int maxn=1e3+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e13+7;


struct is
{
    int x,r;
    bool operator <(const is &c)const
    {
        return x<c.x;
    }
};
int d[maxn],n,m,k;
inline void update(vector<int>&a,vector<int>&b)
{
    priority_queue<is>q;
    for(int i=0;i<b.size();i++)
        d[i] = 0,q.push((is){a[0]+b[i],i});
    vector<int>ans;
    for(int i=1;i<=k;i++)
    {
        if(q.empty())break;
        is x=q.top();
        q.pop();
        ans.push_back(x.x);
        if(d[x.r]+1<a.size())
            q.push((is){a[++d[x.r]]+b[x.r],x.r});
    }
    a=ans;
}
int cmp(int x,int y)
{
    return x>y;
}
struct edge
{
    int from,to,d,nex;
}G[maxn<<2];
int head[maxn],edg;
inline void addedge(int u,int v,int d)
{
    G[++edg]=(edge){u,v,d,head[u]},head[u]=edg;
    G[++edg]=(edge){v,u,d,head[v]},head[v]=edg;
}
int pre[maxn],bccno[maxn];
int dfs_clock,bcc_cnt;
stack<int>s;
vector<int>ans,fuck;
inline int dfs(int u,int fa)
{
    int lowu=++dfs_clock;
    pre[u]=dfs_clock;
    int child=0;
    for(int i=head[u]; i!=-1; i=G[i].nex)
    {
        int v=G[i].to;
        edge e=G[i];
        if(!pre[v])
        {
            s.push(i);
            child++;
            int lowv=dfs(v,u);
            lowu=min(lowu,lowv);
            if(lowv>=pre[u])
            {
                bcc_cnt++;
                fuck.clear();
                while(true)
                {
                    int e=s.top();
                    s.pop();
                    fuck.push_back(G[e].d);
                    if(bccno[G[e].from]!=bcc_cnt)
                        bccno[G[e].from]=bcc_cnt;
                    if(bccno[G[e].to]!=bcc_cnt)
                        bccno[G[e].to]=bcc_cnt;
                    if(G[e].from==u&&G[e].to==v) break;
                }
                if(fuck.size()>1)update(ans,fuck);
            }
        }
        else if(pre[v]<pre[u]&&v!=fa)
        {
            s.push(i);
            lowu=min(lowu,pre[v]);
        }
    }
    return lowu;
}
void init()
{
    dfs_clock=bcc_cnt=edg=0;
    for(int i=0;i<=n;i++)
        head[i]=-1,bccno[i]=0,pre[i]=0;
    ans.clear();
    ans.push_back(0);
}


int main()
{
    int cas=1;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        ll sumsum=0;
        for(int i=1; i<=m; i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            sumsum+=z;
            addedge(x,y,z);
        }
        scanf("%d",&k);
        dfs(1,-1);
        ll out=0,MOD=(1LL<<32);
        printf("Case #%d: ",cas++);

        for(int i=1;i<=k;i++)
        {
            if(i-1>=ans.size())break;
            out+=1LL*(sumsum-ans[i-1])*i;
            out%=MOD;
        }
        printf("%lld\n",out);
    }
    return 0;
}

 

posted @ 2017-07-26 13:52  jhz033  阅读(389)  评论(0编辑  收藏  举报