hdu 4601 Letter Tree

不easy啊。。

一个小错误让我wa死了。找了一个晚上，怎么都找不到

最后是对拍代码找到的错误。发现当步数比較小的时候答案就是对的，比較大的时候就不正确了

想到一定是什么地方越界了。。

。

power[i] = (i64)(power[i - 1] * 26) % mod;

就是这行。。

。

改成  power[i] = ((i64)power[i - 1] * 26) % mod;

就过了。。。

这道题总的来说就是很综合，什么方面都涉及一点，核心部分还是把树转化成序列之后二分边界+RMQ,用dfn来确定边界的这种方法很好，很有意思

事实上另一种方法，就是先从u节点往下走一步。然后在trie里面找，这个时候能够直接确定位置。由于在trie中已经是有序的了

两种都能够，第一种相对来说好实现一点

（hdu如今怎么不会爆栈了。

。）

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;
#ifdef _WIN32
typedef __int64 i64;
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
typedef long long i64;
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define foreach(c,itr)  for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define pf              push_front
#define X               first
#define Y               second
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MC(a,b)		memcpy(a,b,sizeof(b))
#define MAX(a,b)        ((a)>(b)?(a):(b))
#define MIN(a,b)        ((a)<(b)?(a):(b))
#define read            freopen("out.txt","r",stdin)
#define write           freopen("out1.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int mod = 1000000007;
const int maxn = 100111;

struct Node
{
	int now;
	int to;
	int c;
};

struct Trie
{
	int to[26];
	int rank;
}zx[maxn];

int head = 0;
int use;

int get()
{
	use++;
	MM(zx[use].to, -1);
	zx[use].rank = 0;
	return use;
}

int T;
int n, m;
vector<Node>g[maxn];
vector<int>level[maxn];
int dep[maxn];
int dfn[maxn];
int dfv[maxn];
int df;

int power[maxn];
int vis[maxn];
int t[maxn];

int pos[maxn];
int pmod[maxn];
int rankmod[maxn];
int rk;

int c[maxn];

int dfnpos[maxn];
int es[maxn][2];
vector<int>ary;
int st[maxn][20];
int pow2[20];
int lg2[maxn];

void dfs(int now=1,int step=0)
{
	vis[now] = true;
	int to;
	dfn[now] = df;
	dfv[df] = now;
	df++;
	dep[now] = step;
	for (int i = 0; i <(int) g[now].size(); i++)
	{
		to = g[now][i].to;
		if (!vis[to])
		{
			t[to] = now;
			c[to] = g[now][i].c;
			dfs(to,step+1);
		}
	}
}

void make_trie()
{
	use = -1;
	get();
	int trie = head;
	int now = 1;
	pos[now] = trie;
	int to,temp;
	pmod[0] = 0;
	for (int i = 2; i < df; i++)
	{
		to = dfv[i];
		now = t[to];
		trie = pos[now];
		if (zx[trie].to[c[to]] == -1)
		{
			temp = get();
			zx[trie].to[c[to]] = temp;
			pmod[temp] = ((i64)pmod[trie] * 26 + c[to]) % mod;
		}
		pos[to] = zx[trie].to[c[to]];
	}
}

void make_rank(int now = head)
{
	zx[now].rank = rk++;
	for (int i = 0; i < 26; i++)
	{
		if (zx[now].to[i] != -1)
		{
			make_rank(zx[now].to[i]);
		}
	}
}

void sparsetable()
{
	for (int i = 1; i <= n; i++)
	{
		st[i][0] = zx[pos[dfv[ary[i]]]].rank;
	}
	for (int j = 1; j <= lg2[n]; j++)
	{
		for (int i = 1; i <= n; i++)
		{
			if (i + pow2[j - 1] <= n)
			{
				st[i][j] = max(st[i][j - 1], st[i + pow2[j - 1]][j - 1]);
			}
			else
			{
				st[i][j] = st[i][j - 1];
			}
		}
	}
}

int rmq(int l,int r)
{
	return max(st[l][lg2[r - l]], st[r - pow2[lg2[r - l]]][lg2[r - l]]);
}

int find(int x, int step)
{
	int low = dfn[x];
	int up;
	int temp = dfnpos[dfn[x]] + 1;
	step += dep[x];
	int l, r;
	if (temp < es[dep[x]][1])
	{
		up = ary[temp];
	}
	else
	{
		up = inf;
	}
	l = upper_bound (ary.begin() + es[step][0], ary.begin() + es[step][1], low ) - ary.begin();
	r = upper_bound (ary.begin() + es[step][0], ary.begin() + es[step][1], up ) - ary.begin();
	if (l == r)
	{
		return -1;
	}
	int mrank = rmq(l, r);
	int xrank = zx[pos[x]].rank;
	return (((i64)rankmod[mrank] - ((i64)rankmod[xrank] * (i64)power[step - dep[x]])%mod)+mod)%mod;
}

void start()
{
	for (int i = 0; i <= n; i++)
	{
		vis[i] = false;
	}
	t[1] = -1;
	df = 1;
	dfs();
	for (int i = 1; i <= n; i++)
	{
		level[dep[i]].push_back(dfn[i]);
	}
	for (int i = 1; i <= n; i++)
	{
		sort(level[i].begin(), level[i].end());
	}
	ary.clear();
	ary.push_back(0);
	for (int i = 0; i <= n; i++)
	{
		es[i][0] = ary.size();
		for (int j = 0; j < (int)level[i].size(); j++)
		{
			ary.push_back(level[i][j]);
		}
		es[i][1] = ary.size();
	}

	for (int i = 1; i <= n; i++)
	{
		dfnpos[ary[i]] = i;
	}

	make_trie();

	rk = 0;
	make_rank();

	for (int i = 0; i <= use; i++)
	{
		rankmod[zx[i].rank] = pmod[i];
	}

	sparsetable();

}

void destroy()
{
	for (int i = 0; i <= n; i++)
	{
		g[i].clear();
		level[i].clear();
	}
}

int main()
{
	//read;
	//write;
	power[0] = 1;
	for (int i = 1; i < maxn; i++)
	{
		power[i] = ((i64)power[i - 1] * 26) % mod;
	}
	pow2[0] = 1;
	lg2[1] = 0;
	for (int i = 1; i < 20; i++)
	{
		pow2[i] = pow2[i - 1] * 2;
		if(pow2[i]<maxn)
		lg2[pow2[i]] = i;
	}

	for (int i = 3; i < maxn; i++)
	{
		if (!lg2[i])
		{
			lg2[i] = lg2[i - 1];
		}
	}
	cin >> T;
	while (T--)
	{
		cin >> n;
		int x, y;
		char z;
		Node node;
		for (int i = 1; i <= n - 1; i++)
		{
			//cin >> x >> y >> z;
			scanf("%d%d %c", &x, &y, &z);
			node.now = x;
			node.to = y;
			node.c = z - 'a';
			g[node.now].push_back(node);
			swap(node.now, node.to);
			g[node.now].push_back(node);
		}
		start();
		cin >> m;
		for (int i = 1; i <= m; i++)
		{
		//	cin >> x >> y;
			scanf("%d%d", &x, &y);
			if (y == 0)
			{
				printf("0\n");
				continue;
			}
			int ans = find(x, y);
			if (ans == -1)
			{
			//	cout << "IMPOSSIBLE" << endl;
				printf("IMPOSSIBLE\n");
			}
			else
			{
			//	cout << ans << endl;
				printf("%d\n",ans);
			}
		}
		destroy();
	}
	return 0;
}