HDU 5469（2015 ICPC上海网络赛）---Antonidas（树上分治）

题目链接
http://acm.hdu.edu.cn/showproblem.php?pid=5469

Problem Description
Given a tree with N vertices and N−1 edges. Each vertex has a single letter Ci. Given a string S, you are to choose two vertices A and B, and make sure the letters catenated on the shortest path from A to B is exactly S. Now, would you mind telling me whether the path exists?

Input
The first line is an integer T, the number of test cases.
For each case, the first line is an integer N. Following N−1 lines contains two integers a and b, meaning there is an edge connect vertex a and vertex b.
Next line contains a string C, the length of C is exactly N. String C represents the letter on each vertex.
Next line contains a string S.
1≤T≤200, 1≤N≤104, 1≤a,b≤N, a≠b, |C|=N, 1≤|S|≤104. String C and S both only contain lower case letters.

Output
First, please output “Case #k: “, k is the number of test case. See sample output for more detail.
If the path exists, please output “Find”. Otherwise, please output “Impossible”.

Sample Input
2

7
1 2
2 3
2 4
1 5
5 6
6 7
abcdefg
dbaefg

5
1 2
2 3
2 4
4 5
abcxy
yxbac

Sample Output
Case #1: Find
Case #2: Impossible

Source
2015 ACM/ICPC Asia Regional Shanghai Online

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题意：一棵树有n个节点、n-1条边，每个节点上有一个小写英文字母，现有一个只含有英文小写字母的字符串s，问这棵树上是否存在一条路径，串成的字符串和字符串s相同，存在输出“Find” 否则输出“Impossible”；

思路：树上分治，先找到字母为s[0]的节点，然后以它为起点，向下孩子节点或者向上父亲节点搜索，注意剪枝，当向下孩子节点搜索时深度（到叶子节点的最大长度）小于剩余字符串长度时，不必往下搜索。另外，在分治搜索之前，要处理处每个节点的父亲节点和自己的深度；

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
typedef long long LL;
const int maxn=1e4+5;
char C[maxn],S[maxn];
int par[maxn],h[maxn];
bool vis[maxn];
vector<int>q[maxn];
int N,len;

void dfs1(int t,int p)
{
    int tmp=0;
    for(int i=0;i<q[t].size();i++)
    {
        if(q[t][i]!=p){
            par[q[t][i]]=t;
            dfs1(q[t][i],t);
            tmp=max(tmp,h[q[t][i]]);
        }
    }
    h[t]=tmp+1;
}

bool dfs2(int t,int l)
{
    if(l==len) return true;
    for(int i=0;i<q[t].size();i++)
    {
        if(q[t][i]!=par[t]&&len-l<=h[q[t][i]]&&!vis[q[t][i]]&&C[q[t][i]]==S[l]){
            vis[t]=1;
            if(dfs2(q[t][i],l+1)) return true;
            vis[t]=0;
        }
    }
    if(!vis[par[t]]&&C[par[t]]==S[l]){
        vis[t]=1;
        if(dfs2(par[t],l+1)) return true;
        vis[t]=0;
    }
    return false;
}

int main()
{
    int T,Case=1;
    cin>>T;
    while(T--)
    {
        scanf("%d",&N);
        for(int i=0;i<=N;i++)
            q[i].clear();
        for(int i=1;i<N;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            q[u].push_back(v);
            q[v].push_back(u);
        }
        par[1]=0;
        dfs1(1,0);
        scanf("%s%s",C+1,S);
        int flag=0;
        len=strlen(S);
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=N;i++)
        {
            if(C[i]==S[0])
            if(dfs2(i,1)) { flag=1; break; }
        }
        printf("Case #%d: ",Case++);
        if(flag)  puts("Find");
        else      puts("Impossible");
    }
    return 0;
}