HDU-4679 Terrorist’s destroy 树形DP,维护

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4679

　　题意：给一颗树，每个边有一个权值，要你去掉一条边权值w剩下的两颗子树中分别的最长链a,b，使得w*Min(a,b)最小。。

　　说白了就是要枚举每条边，然后在O(1)的时间内求出两颗子树的最长链。因此我们可以考虑用树形DP，首先一遍DFS，对于每个节点维护两个信息，hign[u]：u为根节点的子树的深度，f[u]：u为根节点的子树的最长链。然后还要维护一个hige[i][0]和hige[i][1]，分别表示u为根节点的子树，不包括边 i 的深度和最长链。然后再一遍DFS，根据上一节点的信息递推过去就可以在O(1)的时间内求出来了，总复杂度O(E)。有些细节要考虑，开始把全局变量搞混，wa了几次T^T。。 

//STATUS:C++_AC_875MS_11012KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=100010;
const int INF=0x3f3f3f3f;
const LL MOD=1000000007,STA=8000010;
const LL LNF=1LL<<55;
const double EPS=1e-9;
const double OO=1e50;
const int dx[8]={-1,-1,0,1,1,1,0,-1};
const int dy[8]={0,1,1,1,0,-1,-1,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

struct Edge{
    int u,v,w,id;
}e[N<<1];
int first[N],next[N<<1],hign[N],hige[N<<1][2];
int f[N],maxl1[N],maxr1[N],maxl2[N],maxr2[N],maxlf[N],maxrf[N],d[N],w[N];
int n,mt;
int T,ans,ansid;

void adde(int a,int b,int c,int id)
{
    e[mt].u=a,e[mt].v=b,e[mt].w=c,e[mt].id=id;
    next[mt]=first[a];first[a]=mt++;
    e[mt].u=b,e[mt].v=a,e[mt].w=c,e[mt].id=id;
    next[mt]=first[b];first[b]=mt++;
}

void dfs1(int u,int fa)
{
    int i,j,v,cnt=1,flag=0;
    f[u]=0;
    for(i=first[u];i!=-1;i=next[i]){
        if((v=e[i].v)==fa)continue;
        dfs1(v,u);
        f[u]=Max(f[u],f[v]);
        flag=1;
    }
    if(!flag){f[u]=hign[u]=0;return;}
    for(i=first[u];i!=-1;i=next[i]){
        if((v=e[i].v)==fa)continue;
        w[cnt]=v;
        d[cnt++]=hign[v]+1;
    }
    maxl1[0]=maxr1[cnt]=maxl2[0]=maxr2[cnt]=maxlf[0]=maxrf[cnt]=0;
    for(i=1;i<cnt;i++){
        maxlf[i]=Max(maxlf[i-1],f[w[i]]);
        maxl1[i]=maxl1[i-1],maxl2[i]=maxl2[i-1];
        if(d[i]>maxl1[i])maxl2[i]=maxl1[i],maxl1[i]=d[i];
        else if(d[i]>maxl2[i])maxl2[i]=d[i];
    }
    for(i=cnt-1;i>0;i--){
        maxrf[i]=Max(maxrf[i+1],f[w[i]]);
        maxr1[i]=maxr1[i+1],maxr2[i]=maxr2[i+1];
        if(d[i]>maxr1[i])maxr2[i]=maxr1[i],maxr1[i]=d[i];
        else if(d[i]>maxr2[i])maxr2[i]=d[i];
    }
    for(j=1,i=first[u];i!=-1;i=next[i]){
        if(e[i].v==fa)continue;
        hige[i][0]=Max(maxl1[j-1],maxr1[j+1]);
        hige[i][1]=Max(maxl1[j-1]+maxr1[j+1],
                    maxl1[j-1]+maxl2[j-1],maxr1[j+1]+maxr2[j+1]);
        hige[i][1]=Max(hige[i][1],maxlf[j-1],maxrf[j+1]);
        j++;
    }
    f[u]=Max(f[u],maxr1[1]+maxr2[1]);
    hign[u]=maxr1[1];
}

void dfs2(int u,int fa,int max1,int max2)
{
    int i,j,v,t1,t2,nt;
    for(i=first[u];i!=-1;i=next[i]){
        if((v=e[i].v)==fa)continue;
        t1=Max(hige[i][1],max2,max1+hige[i][0]);
        nt=Max(f[v],t1)*e[i].w;
        if(ans>nt || (ans==nt && e[i].id<ansid)){
            ans=nt;
            ansid=e[i].id;
        }
        t2=Max(max1,hige[i][0])+1;
        dfs2(v,u,t2,Max(t1,t2));
    }
}

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j,a,b,c,ca=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        mem(first,-1);mt=0;
        for(i=1;i<n;i++){
            scanf("%d%d%d",&a,&b,&c);
            adde(a,b,c,i);
        }

        dfs1(1,0);
        ans=INF;
        dfs2(1,0,0,0);

        printf("Case #%d: %d
",ca++,ansid);
    }
    return 0;
}