hdu5469 Antonidas(DFS)

本文介绍了一种在树形结构中进行字符串匹配的方法。通过深度优先搜索算法,判断树上是否存在一条路径使得路径上的节点字母拼接成的目标字符串与给定的字符串相同。该方法考虑了剪枝优化以提高效率。

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思路:直接找到开头的那个结点,然后往子树搜或者父亲结点搜...一个可行的剪枝是如果子树剩下的最大高度小于需要匹配的字符数目的话就肯定没有解了...然后就水过去了...


#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e4+7;
vector<int>e[maxn];
int n,tarlen;
int dep[maxn],vis[maxn],f[maxn];
char s[maxn];
char tar[maxn];
void dfs(int u,int fa)
{
    int d = 0;
//    dep[u]=d;
    f[u]=fa;
    for(int i = 0;i<e[u].size();i++)
    {
        int v = e[u][i];
        if(v!=fa)
        {    
            dfs(v,u);
            d = max(d,dep[v]);
        }
    }
    dep[u]=d+1;
}
bool dfs1(int now,int len)
{
    vis[now]=1;
    if(len==tarlen)
        return true;
    for(int i = 0;i<e[now].size();i++)
    {
        int v = e[now][i];
        if(!vis[v] && f[now]!=v && s[v]==tar[len] && (tarlen-len<=dep[v]))
            if(dfs1(v,len+1))return true;
    }
    int fa = f[now];
    if(!vis[fa] && s[fa]==tar[len])
        if (dfs1(fa,len+1))return true;
    return false;
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i= 0;i<=n;i++)
            e[i].clear();
        memset(dep,0,sizeof(dep));
        for(int i = 1;i<=n-1;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            e[u].push_back(v);
            e[v].push_back(u);
        }
        dfs(1,0);
        scanf("%s",s+1);
        scanf("%s",tar);
        tarlen = strlen(tar);
        bool flag = false;
        for(int i = 1;i<=n;i++)
        {
            if(s[i]==tar[0])
            {
                memset(vis,0,sizeof(vis));
                flag = dfs1(i,1);
                if(flag)
                    break;
            }
        }
        printf("Case #%d: ",cas++);
        if(flag)
            printf("Find\n");
        else
            printf("Impossible\n");
    }
}


Problem Description
Given a tree with  N  vertices and  N1  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  N1  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.
1T200 1N104 1a,bN ab |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
 

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