Gym 100962J Jimi Hendrix (树形DP)

题意：给定一棵树，然后每条边有一个字母，然后给定一行字符串，问你能不能从这棵树上找到，并输出两个端点。

析：树形DP，先进行递归到叶子结点，然后再回溯，在回溯的时候要四个值，一个是正着匹配的长度和端点，一个是反着匹配的长度和端点，

然后一个一个匹配，并不断更新这个长度和端点。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define print(a) printf("%d
", (a))
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e5 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
vector<char> w[maxn];
vector<int> G[maxn];
int ansx, ansy;
struct Node{
    int lans, rans;
    int l, r;
};
Node dp[maxn];
char s[maxn];

bool dfs(int u, int fa){
    for(int i = 0; i < G[u].size(); ++i){
        int v = G[u][i];
        if(v == fa)  continue;
        if(dfs(v, u))  return true;

        int l = s[dp[v].l+1] == w[u][i] ? dp[v].l+1 : dp[v].l;
        int r = s[m-dp[v].r] == w[u][i] ? dp[v].r+1 : dp[v].r;
        if(l + dp[u].r >= m){
            ansx = dp[v].lans;
            ansy = dp[u].rans;
            return true;
        }
        else if(r + dp[u].l >= m){
            ansx = dp[u].lans;
            ansy = dp[v].rans;
            return true;
        }
        if(l > dp[u].l){
            dp[u].l = l;
            dp[u].lans = dp[v].lans;
        }
        if(r > dp[u].r){
            dp[u].r = r;
            dp[u].rans = dp[v].rans;
        }
    }
    return false;
}

int main(){
    while(scanf("%d %d", &n, &m) == 2){
        for(int i = 1; i <= n; ++i)  G[i].clear(), w[i].clear();
        int u, v;
        char ch;
        for(int i = 1; i < n; ++i){
            scanf("%d %d %c", &u, &v, &ch);
            dp[i].l = dp[i].r = 0;
            dp[i].lans = dp[i].rans = i;
            w[u].push_back(ch);
            w[v].push_back(ch);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        dp[n].l = dp[n].r = 0;
        dp[n].lans = dp[n].rans = n;

        scanf("%s", s+1);
        ansx = ansy = -1;
        dfs(1, -1);
        printf("%d %d
", ansx, ansy);
    }
    return 0;
}