HDU - 6041：I Curse Myself（Tarjan求环&K路归并）

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

Please calculate (∑ k=1 K k⋅V(k))mod2 32  (∑k=1Kk⋅V(k))mod232 .

InputThe input contains multiple test cases.

For each test case, the first line contains two positive integers n,m n,m (2≤n≤1000,n−1≤m≤2n−3) (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 m lines contains three positive integers x,y,z x,y,z (1≤x,y≤n,1≤z≤10 6 ) (1≤x,y≤n,1≤z≤106) , meaning an edge weighted z z between node x x and node y y . There does not exist multi-edge or self-loop in this graph.

The last line contains a positive integer K K (1≤K≤10 5 ) (1≤K≤105) .OutputFor each test case, output " Case #x x : y y " in one line (without quotes), where x x indicates the case number starting from 1 1 and y 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

题意：给一个仙人掌，求前K小的生成树的值之和。

思路：由于是仙人掌，即每个环去掉一条边才能得到生成树，那么就算每个环贡献一个数，求前K大。

每个集合贡献一个，求前K大我们可以用K路归并。 从左到右两两合并，因为最后只要K个数，那么K以后的就不用保存了，所以每次合并最多保留K个数，则合并完的复杂度为O(NK)；  

K路归并：当假设答案集合为A，新集合为B。那么对于B的每个数i，先把bi+a0放入大根堆里，然后开始操作：每次取队首X放入答案集合里，然后把对应的bi+a(j+1)又插入堆里，因为它是关于bi的最大的小于X的数，这样一直操作可以保证是最大的K个。 （和“超级钢琴”那题一样，如果用了一个数X，我才把它分裂后的数（小于X）拿来考虑）。

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=400010;
int Laxt[maxn],Next[maxn],To[maxn],Len[maxn],q[maxn],cnt;
int dfn[maxn],low[maxn],times,scc[maxn],scc_cnt,head,K;
vector<int>G[maxn];
struct in{
    int w,a,b;
    in(){}
    in(int ww,int aa,int bb):w(ww),a(aa),b(bb){}
    friend bool operator <(in x,in y){ return x.w<y.w;}
};
void add(int u,int v,int w){
    Next[++cnt]=Laxt[u]; Laxt[u]=cnt; To[cnt]=v; Len[cnt]=w;
}
void tarjan(int u,int fa)
{
    dfn[u]=low[u]=++times;
    q[++head]=u;
    for(int i=Laxt[u];i;i=Next[i]){
        if(To[i]==fa) continue;
        if(!dfn[To[i]]) {
              tarjan(To[i],u);
              low[u]=min(low[u],low[To[i]]);
        }
        else low[u]=min(low[u],dfn[To[i]]);
    }
    if(low[u]==dfn[u]){
        scc_cnt++;
        while(true){
            int x=q[head--];
            scc[x]=scc_cnt;
            if(x==u) break;
        }
    }
}
void merge(vector<int>&a,vector<int>b){
    vector<int>res;
    priority_queue<in>q;
    for(int i=0;i<b.size();i++)
      q.push(in(a[0]+b[i],0,i));
    while(res.size()<K&&!q.empty()){
        in it=q.top(); q.pop();
        res.push_back(it.w);
        if(it.a+1<a.size()){
            it.a++;     //当前数没了就把当前数的次大值拉进来，ok。
            q.push(in(a[it.a]+b[it.b],it.a,it.b));
        }
    }
    a=res;
}
int main()
{
    int N,M,u,v,w,C=0,sum; unsigned int res;
    while(~scanf("%d%d",&N,&M)){
        rep(i,1,N) Laxt[i]=dfn[i]=scc[i]=low[i]=0;
        cnt=times=scc_cnt=head=0; sum=0; res=0;
        rep(i,1,M) {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w); add(v,u,w); sum+=w;
        }
        tarjan(1,0);
        rep(i,1,scc_cnt) G[i].clear();
        rep(i,1,N)
          for(int j=Laxt[i];j;j=Next[j])
            if(To[j]>i&&scc[i]==scc[To[j]])
              G[scc[i]].push_back(Len[j]);
        scanf("%d",&K);
        vector<int>ans; ans.push_back(0);
        rep(i,1,scc_cnt){
            if(G[i].size()>0) merge(ans,G[i]);
        }
        for(int i=0;i<ans.size();i++) res+=(i+1)*(sum-ans[i]);
        printf("Case #%d: %u\n",++C,res);
    }
    return 0;
}

无向图tarjan：按边存。(想象一下有的点存在于多个环里)

void tarjan(int u,int fa)
{
    dfn[u]=low[u]=++times; vis[u]=1;
    for(int i=Laxt[u];i;i=Next[i]){
        int v=To[i];
        if(!F[i]) continue;
        F[i]=F[i^1]=0;
        if(!vis[v]){
            q[++head]=i;
            tarjan(v,u);
            low[u]=min(low[u],low[v]);
            if(dfn[u]<=low[v])
            {
                int k,fcy=0; scc_cnt++;
                do
                {
                    k=q[head--];
                    fcy++;
                    scc[k]=scc_cnt;
                } while (k!=i);
                if(fcy>1) ans1++,ans2=(ll)ans2*(fcy+1)%Mod;;
            }
        }
        else if(dfn[v]<dfn[u]){
            q[++head]=i;
            low[u]=min(low[u],dfn[v]);
        }
    }
}

 

 

 

裸题K路归并：POJ2442，M个集合，每个集合大小为N，每个集合选一个数加起来，问前N小。

#include<bits/stdc++.h>
using namespace std;
const int maxn=2010;
struct in{
    int x,pos;
    in(){}
    in(int xx,int pp):x(xx),pos(pp){}
    friend bool operator <(in w,in v){return w.x>v.x;}
};
void merge(vector<int>&G,int N)
{
    vector<int>res; priority_queue<in>q;
    for(int i=1,x;i<=N;i++){
        scanf("%d",&x);
        q.push(in(G[0]+x,0));
    }
    while(!q.empty()&&res.size()<N){
        in w=q.top(); q.pop();
        res.push_back(w.x);
        if(w.pos+1<G.size())
            q.push(in(w.x+G[w.pos+1]-G[w.pos],w.pos+1));
    }
    G=res;
}
int main()
{
    int T,M,N;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&M,&N);
        vector<int>G ;
        G.push_back(0);
        for(int i=1;i<=M;i++) merge(G,N);
        for(int i=0;i<N;i++) printf("%d ",G[i]);
        puts("");
    }
    return 0;
}