HDU-3660 Alice and Bob's Trip 树形dp

题意：有一棵树，Alice和Bob两个人要从树根走到叶子。Alice想要路径尽量短，Bob想要路径尽量长，且路径长度一定要在[L,R]区间范围内。从根节点开始Bob和Alice轮流选择走那条路，问Bob能选到的最长路径是什么？

解法：这道题一看到就想Bob从儿子中选最长的，Alice从儿子中选最短的就可以了，码了之后就过了。。。但是之后想想觉得有点不对劲，因为Bob想要尽量长，要是他这回合选了最长的儿子，那么下一回合就是Alice选了，那么Alice会不会选一条对Bob不利的路，使得其实Bob在上一回合选一条次长的而不是最长的反而会更划算。（如果按照这个思路想的话，那Bob就应该从儿子中选一个最短的最长来避免最保证最坏情况的发生）。关于这个疑问蒟蒻博主还现在没想明白。

另外，这道题对时间要求极高，博主需要用到输入挂才顺利通过。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int N=5e5+10;
const int INF=0x3f3f3f3f;
int n,l,r,indeg[N];

//inline int read() {
//    int x=0, f=1; register char ch=getchar();
//    for (; ch<'0' || ch>'9'; ch=getchar()) if (ch=='-') f=-1;
//    for (; ch>='0' && ch<='9'; ch=getchar()) x=x*10+ch-'0';
//    return x*f;
//}
#define reg register
#define fok (ch!=EOF)
#define sep (ch==' '||ch=='
'||ch=='	')
#define dsep !isdigit(ch)
#define neq(a,b) ((a)-(b)>1e-6)
struct FastIO{
    char rbuf[1000002],wbuf[1000002],b,*p1,*p2;
    int rs,ws,S;
    FastIO(){p1=rbuf,p2=wbuf,S=1000000,rs=1000000,ws=-1,b=1;}
    inline void go(){fwrite(wbuf,1,ws+1,stdout)?ws=-1:0;}
    inline char getch(){
        return rs==S&&
        (p1=rbuf,rs=-1,(S=fread(rbuf,1,S+1,stdin)-1)==-1)?
        (b=0,EOF):(++rs,*p1++);
    }
    inline void putch(int x){
        ws==1000000&&
        (p2=wbuf,ws=-1,fwrite(wbuf,1,1000001,stdout)),++ws,*p2++=x;
    }
    inline void puts(char str[]){
        for(reg int i=0;str[i]!='';)putch(str[i]),++i;
    }
    inline void getline(string& s){
        for(reg char ch;(ch=getch())!='
'&&fok;)s+=ch;
    }
    inline FastIO& operator>>(int& x){
        x=0;reg char f=0,ch=getch();
        while(dsep&&fok)f|=(ch=='-'),ch=getch();
        while(isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getch();
        return x=f?-x:x,*this;
    }
    inline FastIO& operator>>(long long& x){
        x=0;reg char f=0,ch=getch();
        while(dsep&&fok)f|=(ch=='-'),ch=getch();
        while(isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getch();
        return x=f?-x:x,*this;
    }
    inline FastIO& operator>>(char& ch){return ch=getch(),*this;}
    inline FastIO& operator>>(string& s){
        reg char ch=getch();
        while(sep&&fok)ch=getch();
        while(!sep&&fok)s+=ch,ch=getch();
        return *this;
    }
    inline FastIO& operator>>(double& x){
        x=0;reg char f=0,ch=getch();
        double d=0.1;
        while(dsep&&fok)f|=(ch=='-'),ch=getch();
        while(isdigit(ch))x=x*10+(ch^48),ch=getch();
        if(ch=='.')
            while(isdigit(ch=getch()))x+=d*(ch^48),d*=0.1;
        return x=f?-x:x,*this;
    }
    inline FastIO& operator<<(int x){
        reg char ch[10],i=-1;
        !x?(putch('0'),0):0,
        x<0?(putch('-'),x=-x):0;
        while(x)ch[++i]=x%10+48,x/=10;
        while(~i)putch(ch[i]),--i;
        return *this;
    }
    inline FastIO& operator<<(long long x){
        reg char ch[20],i=-1;
        !x?(putch('0'),0):0,
        x<0?(putch('-'),x=-x):0;
        while(x)ch[++i]=x%10+48,x/=10;
        while(~i)putch(ch[i]),--i;
        return *this;
    }
    inline FastIO& operator<<(char ch){return putch(ch),*this;}
    inline FastIO& operator<<(char str[]){return puts(str),*this;}
    inline FastIO& operator<<(string s){
        reg int nn=s.size();
        for(reg int i=0;i<nn;++i)putch(s[i]);
        return *this;
    }
    inline FastIO& operator<<(double x){
        int y=int(x);
        *this<<y;
        x-=y;
        while(neq(x,int(x)))x*=10;
        x?*this<<'.'<<int(x):0;
        return *this;
    }
    inline operator bool(){return b;}
};
FastIO io;  //输入挂 

int cnt=1,head[N],nxt[N<<1],to[N<<1],Len[N<<1];
void add_edge(int x,int y,int z) {
    nxt[++cnt]=head[x]; to[cnt]=y; Len[cnt]=z; head[x]=cnt;
}

int dp[N];
void dfs(int x,int dep,int len,int fa) {
    if (indeg[x]==1 && x!=1) {  //Leaf
        if (l<=len && len<=r) dp[x]=0; else dp[x]=-1;
        return;
    }
    if (dep%2==1) dp[x]=0; else dp[x]=INF;
    for (int i=head[x];i;i=nxt[i]) {
        int y=to[i];
        if (y==fa) continue;
        dfs(y,dep+1,len+Len[i],x);
        if (dp[y]==-1) continue;
        if (dep%2==1) {  //Bob
            dp[x]=max(dp[x],dp[y]+Len[i]);
        } else {  //Alice
            dp[x]=min(dp[x],dp[y]+Len[i]);
        }
    }
    if (dp[x]==0 || dp[x]==INF) dp[x]=-1;
}

int main()
{
    while (io>>n) {
        io>>l>>r;
        cnt=1; for (int i=1;i<=n;i++) head[i]=0,indeg[i]=0;
        for (int i=1;i<n;i++) {
            int x,y,z; io>>x>>y>>z;
            x++; y++; indeg[x]++; indeg[y]++;
            add_edge(x,y,z); add_edge(y,x,z);
        }
        dfs(1,1,0,0);
        if (dp[1]==-1) puts("Oh, my god!"); else printf("%d
",dp[1]);
    }
    return 0;
}