【搜索剪枝】HDU 5469 Antonidas

通道

题意：给出1字母树，询问一字符串是否出现在该树中

思路：直接搜索剪枝，有人点分治？写了几发都T了。。有人会了教我？

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

struct Edge {
    int v, nxt;
    Edge () {
        
    }
    Edge (int _v, int _n) {
        v = _v, nxt = _n;
    }
};

const int MAX_N = 10007;

int head[MAX_N], n, len, ecnt, Tim;
char a[MAX_N], b[MAX_N];
int dis[MAX_N], f[MAX_N], vis[MAX_N];
Edge G[MAX_N << 1];

void init() {
    ecnt = Tim = 0;
    for (int i = 0; i <= n; ++i) head[i] = -1;
}

void add(int u, int v) {
    G[ecnt] = Edge(v, head[u]);
    head[u] = ecnt++;
}

void dfs(int u, int fa) {
    for (int i = head[u]; ~i; i = G[i].nxt) {
        int v = G[i].v;
        if (v == fa) continue;
        dfs(v, u);
        dis[u] = max(dis[u], dis[v] + 1);
        f[v] = u;
    }
    if (dis[u] == -1) dis[u] = 1;
}

bool Find(int u, int dep, int r) {
    vis[u] = Tim;
    if (dep > len) return true;
    if(dis[u] > r)
        for(int i = head[u]; ~i; i = G[i].nxt) {
            int v = G[i].v;
            if(a[v] == b[dep] && Tim != vis[v]) 
                if(Find(v, dep + 1, r - 1)) return true;
        }
    else {
        if(a[f[u]] == b[dep] &&  Tim != vis[f[u]]) 
            return Find(f[u], dep + 1, r - 1);
    }
    return false;
}

int main() {
    int T; scanf("%d", &T);
    int cas = 0;
    while(T--) {
        scanf("%d", &n);
        init();
        int u, v;
        for (int i = 1; i < n; ++i) {
            scanf("%d%d", &u, &v);
            add(u, v), add(v, u);
        }
        scanf("%s%s",a + 1, b + 1);  
        memset(dis, -1, sizeof dis);
        memset(vis, 0, sizeof vis);
        dfs(1, -1); 
        bool f = 0;
        len = strlen(b + 1);
        for(int i = 1; i <= n; ++i)
            if(a[i] == b[1]) {
                ++Tim;
                if(Find(i, 2, len - 1)) {
                    printf("Case #%d: Find\n", ++cas);
                    f = true;
                    break;    
                }
            }    
        if(!f) printf("Case #%d: Impossible\n", ++cas);
    }
    return 0;
}