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  • (贪心+倍增求LCA) hdu 4912

    Paths on the tree

    Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 1097    Accepted Submission(s): 366


    Problem Description
    bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n.

    There are m paths on the tree. bobo would like to pick some paths while any two paths do not share common vertices.

    Find the maximum number of paths bobo can pick.
     
    Input
    The input consists of several tests. For each tests:

    The first line contains n,m (1≤n,m≤105). Each of the following (n - 1) lines contain 2 integers ai,bi denoting an edge between vertices ai and bi (1≤ai,bi≤n). Each of the following m lines contain 2 integers ui,vi denoting a path between vertices ui and vi (1≤ui,vi≤n).
     
    Output
    For each tests:

    A single integer, the maximum number of paths.
     
    Sample Input
    3 2 1 2 1 3 1 2 1 3 7 3 1 2 1 3 2 4 2 5 3 6 3 7 2 3 4 5 6 7
     
    Sample Output
    1 2
     
    Author
    Xiaoxu Guo (ftiasch)
     
    Source
     
     
    从最深的公共祖先开始选择,选完之后,他的所有子节点都不可能再选,不然就重复了
     
    用倍增思想求解即可
    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<cstdlib>
    #include<string>
    #include<algorithm>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<map>
    #include<set>
    using namespace std;
    vector<int> e[100005];
    struct node
    {
        int x,y,z;
    }mp[100005];
    int n,m,vis[100005],deep[100005],f[100005][30],mark[100005];
    void init()
    {
        for(int i=1;i<=n;i++)
        {
            vis[i]=0;
            deep[i]=0;
            mark[i]=0;
            e[i].clear();
        }
        memset(f,0,sizeof(f));
    }
    void dfs(int u)
    {
        vis[u]=1;
        for(int i=0;i<e[u].size();i++)
        {
            int v=e[u][i];
            if(v==f[u][0])
                continue;
            if(!vis[v])
            {
                deep[v]=deep[u]+1;
                f[v][0]=u;
                dfs(v);
            }
        }
    }
    void bz()
    {
        for(int i=1;(1<<i)<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                f[j][i]=f[f[j][i-1]][i-1];
            }
        }
    }
    bool cmp(node a,node b)
    {
        return deep[a.z]>deep[b.z];
    }
    int LCA(int x,int y)
    {
        int i,j;
        if(deep[x]<deep[y])
            swap(x,y);
        for(i=0;(1<<i)<=deep[x];i++);
        i--;
        for(j=i;j>=0;j--)
        {
            if(deep[x]-(1<<j)>=deep[y])
            {
                x=f[x][j];
            }
        }
        if(x==y)
            return y;
        for(j=i;j>=0;j--)
        {
            if(f[x][j]!=f[y][j])
            {
                x=f[x][j];
                y=f[y][j];
            }
        }
        return f[x][0];
    }
    void col(int u)
    {
        mark[u]=1;
        for(int i=0;i<e[u].size();i++)
        {
            int v=e[u][i];
            if(deep[v]==deep[u]+1&&!mark[v])
                col(v);
        }
    }
    int main()
    {
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            init();
            for(int i=1;i<n;i++)
            {
                int x,y;
                scanf("%d%d",&x,&y);
                e[x].push_back(y);
                e[y].push_back(x);
            }
            f[1][0]=1;
            deep[1]=1;
            dfs(1);
            bz();
            for(int i=0;i<m;i++)
            {
                int x,y;
                scanf("%d%d",&x,&y);
                mp[i].x=x,mp[i].y=y,mp[i].z=LCA(x,y);
            }
            int ans=0;
            sort(mp,mp+m,cmp);
            for(int i=0;i<m;i++)
            {
                if(!mark[mp[i].x]&&!mark[mp[i].y])
                {
                    ans++;
                    col(mp[i].z);
                }
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/water-full/p/4502642.html
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