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-8using 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);    }}