题目链接：zoj 3817 Chinese Knot

题目大意：给出四个字符串，对应着同心结的四条边，现在给定一个目标串，可以从任意节点开始移动，问是否可以匹配目标串。

解题思路：用hash将四个字符串的正序和逆序处理出来，然后dfs枚举，每次保留起始位置和移动方向即可。

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

using namespace std;
typedef unsigned long long ull;
typedef pair<int, int> pii;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)

const int maxn = 100005;
const ull X = 107;

char str[maxn];
int N, M;
vector<pii> ans;
ull G[12][maxn], H[maxn], T[maxn];

void init () {
    scanf("%d%d", &N, &M);
    for (int i = 0; i < 4; i++) {
        scanf("%s", str+1);

        G[lson(i)][N+1] = G[rson(i)][N+1] = 0;
        for (int j = N; j >= 1; j--) {
            G[lson(i)][j] = G[lson(i)][j+1] * X + (str[j] - ‘a‘);
            G[rson(i)][j] = G[rson(i)][j+1] * X + (str[N-j+1] - ‘a‘);
        }
    }

    scanf("%s", str+1);
    H[M+1] = 0;
    for (int j = M; j >= 1; j--)
        H[j] = H[j+1] * X + (str[j] - ‘a‘);
}

bool dfs (int k, int pos, int dir, int L) {
    if (L == 0)
        return true;

    int len = min(N - pos + 1, L);
    if (G[(k<<1)|dir][pos] - G[(k<<1)|dir][pos+len] * T[len] == H[M-L+1] - H[M-L+len+1] * T[len]) {
        ans.push_back(make_pair(k * N + (dir ? N + 1 - pos : pos), dir));
        L -= len;

        if (L == 0)
            return true;

        for (int i = 0; i < 4; i++) {
            for (int d = 0; d < 2; d++) {
                if (k == i && (d^1) == dir)
                    continue;

                if (dfs(i, 1, d, L))
                    return true;
            }
        }

        ans.pop_back();
    }

    return false;
}

bool solve () {
    for (int i = 0; i < 4; i++) {
        for (int pos = 1; pos <= N; pos++) {
            for (int d = 0; d < 2; d++) {
                ans.clear();
                if (dfs(i, pos, d, M))
                    return true;
            }
        }
    }
    return false;
}

int main () {
    int cas;
    scanf("%d", &cas);

    T[0] = 1;
    for (int i = 1; i < maxn; i++)
        T[i] = T[i-1] * X;

    while (cas--) {
        init();
        if (solve()) {

            int p = 0;
            for (int k = 0; k < ans.size(); k++) {
                int u = (ans[k].first - 1) / N + 1, dir = ans[k].second;

                if (dir == 0) {
                    for (int i = ans[k].first; i <= u * N && p < M; i++, p++) {
                        if (p) printf(" ");
                        printf("%d", i);
                    }
                } else {
                    for (int i = ans[k].first; i > (u-1) * N && p < M; i--, p++) {
                        if (p) printf(" ");
                        printf("%d", i);
                    }
                }
            }
            printf("\n");
        } else
            printf("No solution!\n");
    }
    return 0;
}

zoj 3817 Chinese Knot(hash+暴力)