hdu 6041 I Curse Myself 无向图找环+优先队列

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 32.

 

 

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 #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

 

#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;
}