[USACO 6.3.2] Cryptcowgraphy

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題目大意

　　現在知道一個字串"Begin the Escape execution at the Break of Dawn",在其中按順序插入‘C‘,‘O‘,‘W‘.‘C‘和‘O‘之間的部分會和‘O‘‘W‘之間的部分調換.

　　給一個字串,問此字串是不是又上述變換形成的並輸出經曆多少次變換.

題解

　　這是黑書<<演算法藝術與資訊學競賽>>中的一道搜尋題.

　　很明顯,如果樸素的演算法是不可能AC的.因此,我們要加入一些最佳化.

　　　　1.搜尋順序: 聽說加入這個剪枝也是會快不少的,我們的迴圈順序應該是‘O‘->‘C‘->‘W‘,反正就是先中間後兩邊.

　　　　2.可行性剪枝: 對於固定部分,即此後無論怎麼搜都不變的,也就是"COW"其中一個字元到下一個"COW"字元的中間部分,我們可以在原串中先行匹配,如果不匹配那麼可以進行剪枝.

　　　　3.重複狀態: 顧名思義判重,這得用傳說中的ELFhash,我至今沒搞懂是什麼意思,不過沒關係,能用就行.

代碼

 

/*TASK:cryptcowLANG:C++*/#include <cstdio>#include <cstring>using namespace std;const char *S = "Begin the Escape execution at the Break of Dawn";const int lens = strlen(S);const int MODU = 999991;char buf[100];int n, node;bool hash[MODU];bool check(){    int cnt[256], k = 0;    memset(cnt, 0, sizeof(cnt));    for (int i = 0; i < lens; ++i)        cnt[S[i]]++;    for (int i = 0; i < n; ++i)        if (buf[i] != ‘C‘ && buf[i] != ‘O‘ && buf[i] != ‘W‘)            cnt[buf[i]]--;        else k++;    for (int i = 0; i < 256; ++i)        if (cnt[i]) return false;    return k % 3 == 0 && lens == n - k;}bool match(int k){    if (k == n || buf[k] == ‘C‘ || buf[k] == ‘O‘ || buf[k] == ‘W‘) return true;    int f[200];    f[0] = f[1] = 0;    for (int i = 1; i < n - k; ++i)    {        int j = f[i];        if (buf[k + i] == buf[k + j]) f[i + 1] = j + 1;        else f[i + 1] = 0;    }    int j = 0;    for (int i = 0; i < lens; ++i)    {        while (j != 0 && S[i] != buf[k + j]) j = f[j];        if (S[i] == buf[k + j]) j++;        if (k + j == n || buf[k + j] == ‘C‘ || buf[k + j] == ‘O‘ || buf[k + j] == ‘W‘) return true;    }    return false;}bool judge(){    for (int i = 0; i < n; ++i)    {        if (buf[i] == ‘W‘ || buf[i] == ‘C‘ || buf[i] == ‘O‘)        {            if (buf[i] == ‘C‘) break;            if (buf[i] == ‘O‘ || buf[i] == ‘W‘) return false;        }        else if (buf[i] != S[i]) return false;    }    for (int i = 0; i < n; ++i)        if (buf[i] == ‘C‘ || buf[i] == ‘O‘ || buf[i] == ‘W‘)            if (!match(i + 1)) return false;    return true;}int ELFhash(char *str){    unsigned int h = 0, g;    while (*str)    {        h = (h << 4) + (*str++);        if (g = h & 0xf0000000l)        {            h ^= g >> 24;        }        h &= ~g;    }    return h % MODU;}bool dfs(int dep){    unsigned int h = ELFhash(buf);    if (hash[h]) return false;    hash[h] = true;    if (!strcmp(buf, S))    {        printf("1 %d\n", dep);        return true;    }    char now[100];    int len = n;    memcpy(now, buf, sizeof(buf));    for (int j = 0; j < len; ++j) if (now[j] == ‘O‘)        for (int i = 0; i < j; ++i) if (now[i] == ‘C‘)            for (int k = len - 1; k > j; --k) if (now[k] == ‘W‘)            {                n = 0;                for (int lv = 0; lv < i; ++lv)                    buf[n++] = now[lv];                for (int lv = j + 1; lv < k; ++lv)                    buf[n++] = now[lv];                for (int lv = i + 1; lv < j; ++lv)                    buf[n++] = now[lv];                for (int lv = k + 1; lv < len; ++lv)                    buf[n++] = now[lv];                buf[n] = ‘\0‘;                if (judge() && dfs(dep + 1)) return true;            }    return false;}int main(){    freopen("cryptcow.in", "r", stdin);    freopen("cryptcow.out", "w", stdout);    fgets(buf, 100, stdin);    n = strlen(buf) - 1;    buf[n] = ‘\0‘;    node = 0;    memset(hash, false, sizeof(hash));    if (!check() || !judge() || !dfs(0)) printf("0 0\n");    return 0;}

 

[USACO 6.3.2] Cryptcowgraphy