不容易啊。。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，发现当步数比较小的时候答案就是对的，比较大的时候就不对了
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非常综合，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来确定边界的这个方法非常好，很有意思
其实还有一种方法，就是先从u节点往下走一步，然后在trie里面找，这个时候可以直接确定位置，因为在trie中已经是有序的了
两种都可以，第一种相对来说好实现一点
（hdu现在怎么不会爆栈了。。）
//#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;
}
hdu 4601 Letter Tree 2013多校1-2,布布扣,bubuko.com
hdu 4601 Letter Tree 2013多校1-2
原文地址：http://blog.csdn.net/zz_1215/article/details/38024563
