HDU 3660 Alice and Bob's Trip

树形dp，这道题如果选G++的话，只输入都会超时。我是C++ 1900ms + 飘过的。。。但是输入优化后就快了很多了，1100ms左右。dfs按层次求最值就行了，差不多也算是博弈吧，到bob取的时候要选尽量大的分支（满足条件L和R之间的情况下），反之要alice选尽量小的分支。然后一遍dfs就可以了，时间复杂度为O(n)。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<fstream>
#include<sstream>
#include<vector>
#include<string>
#include<cstdio>
#include<bitset>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
#define MP make_pair
#define eps 1e-8
using namespace std;

const int N = 500005;
const int INF = 1e9 + 7;

bool read(int &x)
{
    char c;
    while(c=getchar())
    {
        if(c == EOF) return 0;
        if(c<='9'&&c>='0')break;
    }
    x=c-'0';
    while(c=getchar())
    {
        if(c == EOF) return 0;
        else  if(c>'9'||c<'0')break;
        else x=x*10+c-'0';
    }
    return 1;
}

struct Edge
{
    int u, v, c;
} E[N << 1];

int fir[N], next[N << 1], tot, l, r;

void Add_Edge(int u, int v, int c)
{
    E[tot].u = u, E[tot].v = v, E[tot].c = c;
    next[tot] = fir[u], fir[u] = tot ++;
}

int dfs(int u, int fa, int c, bool lvl)
{
    int v, ret = INF, tmp;
    if(lvl) ret = 0;
    for(int i = fir[u]; ~i; i= next[i])
    {
        v = E[i].v;
        if(v != fa)
        {
            tmp = dfs(v, u, c + E[i].c, !lvl) + E[i].c;
            if((tmp + c >= l) && (tmp + c <= r))
            {
                if(lvl) ret = max(ret, tmp);
                else ret = min(ret, tmp);
            }
        }
    }
    if(ret == INF) ret = 0;
    return ret;
}

int main()
{
    int n, u, v, c, ans;
    while(read(n))
    {
        read(l);read(r);
        CLR(fir, -1);
        tot = 0;
        for(int i = 1; i < n; i ++)
        {
            read(u);read(v);read(c);
            Add_Edge(u, v, c);
            Add_Edge(v, u, c);
        }
        ans = -1;
        for(int i = fir[0]; ~i; i = next[i])
        {
            v = E[i].v;
            int tmp = dfs(v, 0, E[i].c, 0) + E[i].c;
            if(tmp >= l && tmp <= r)
            {
                ans = max(ans, tmp);
            }
        }
        if(ans == -1) puts("Oh, my god!");
        else printf("%d\n", ans);
    }
}