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  • HDU 2196 Computer 树形DP

    求每个节点到距离他最远节点的距离。

    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<algorithm>
    #include<iostream>
    #include<sstream>
    #include<cmath>
    #include<climits>
    #include<string>
    #include<map>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<set>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<int,int> pii;
    #define pb(a) push_back(a)
    #define INF 0x1f1f1f1f
    #define lson idx<<1,l,mid
    #define rson idx<<1|1,mid+1,r
    #define PI  3.1415926535898
    template<class T> T min(const T& a,const T& b,const T& c) {
        return min(min(a,b),min(a,c));
    }
    template<class T> T max(const T& a,const T& b,const T& c) {
        return max(max(a,b),max(a,c));
    }
    void debug() {
    #ifdef ONLINE_JUDGE
    #else
    
        freopen("d:\in.txt","r",stdin);
       // freopen("d:\out1.txt","w",stdout);
    #endif
    }
    int getch() {
        int ch;
        while((ch=getchar())!=EOF) {
            if(ch!=' '&&ch!='
    ')return ch;
        }
        return EOF;
    }
    
    const int maxn=10100;
    
    struct EDGE
    {
        int u,v,w;
        EDGE(){}
        EDGE(int a,int b,int c):u(a),v(b),w(c){}
    };
    vector<EDGE> e;
    vector<int> g[maxn];
    
    ll f1[maxn],f2[maxn];
    int son1[maxn],son2[maxn];
    
    
    ll StepOne(int x,int fa)
    {
        f1[x]=0;f2[x]=0;
        son1[x]=-1;son2[x]=-1;
        for(int i=0;i<g[x].size();i++)
        {
            int v=e[g[x][i]].v,w=e[g[x][i]].w;
            if(v!=fa)
            {
                ll res=StepOne(v,x);
    
                if(res+w>=f1[x])
                {
                    f2[x]=f1[x];
                    son2[x]=son1[x];
    
                    f1[x]=res+w;
                    son1[x]=v;
                }
                else if(res+w>=f2[x])
                {
                    f2[x]=res+w;
                    son2[x]=v;
                }
            }
        }
        return f1[x];
    }
    
    int StepTwo(int x,int fa,int ei)
    {
        if(fa!=-1)
        {
            if(son1[fa]!=x)
            {
                if(f1[fa]+e[ei].w>=f1[x])
                {
                    f2[x]=f1[x];
                    son2[x]=son1[x];
    
                    f1[x]=f1[fa]+e[ei].w;
                    son1[x]=fa;
                }else if(f1[fa]+e[ei].w>=f2[x])
                {
                    f2[x]=f1[fa]+e[ei].w;
                    son2[x]=fa;
                }
            }else
            {
                if(f2[fa]+e[ei].w>=f1[x])
                {
                    f2[x]=f1[x];
                    son2[x]=son1[x];
    
                    f1[x]=f2[fa]+e[ei].w;
                    son1[x]=fa;
                }else if(f2[fa]+e[ei].w>=f2[x])
                {
                    f2[x]=f2[fa]+e[ei].w;
                    son2[x]=fa;
                }
            }
        }
    
        for(int i=0;i<g[x].size();i++)
        {
            int v=e[g[x][i]].v;
            if(v!=fa)
            {
                StepTwo(v,x,g[x][i]);
            }
        }
        return 0;
    }
    void read(int n)
    {
        for(int i=1;i<=n;i++)
            g[i].clear();
        e.clear();
        for(int i=2;i<=n;i++)
        {
            int v,w;
            scanf("%d%d",&v,&w);
            e.push_back(EDGE(i,v,w));
            g[i].push_back(e.size()-1);
    
            e.push_back(EDGE(v,i,w));
            g[v].push_back(e.size()-1);
        }
    }
    int main()
    {
       // freopen("in.txt","r",stdin);
        int n;
        while(scanf("%d",&n)!=EOF)
        {
            read(n);
            StepOne(1,-1);
            StepTwo(1,-1,-1);
            for(int i=1;i<=n;i++)
                printf("%I64d
    ",f1[i]);
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/BMan/p/3419613.html
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