Light 1039 - A Toy Company (bfs)

题意

给一个初始串和目标串，一次只能变换相邻的字母，问最少变换几次，有一些禁止变换的排列。

思路

因为总共才3个字母，所以可以直接用数字表示一个单词。

然后处理一下禁止的单词，丢set里。

然后就开始bfs。

这题的思路很简单，不过写得很蛋疼。。不知道为什么。把状态push到队列的时候的写法换了很多种才写出来。。

这题也可以建图跑最短路。不过感觉跑不跑都差不多，因为边权都是1.

代码

#include <stack>#include <cstdio>#include <list>#include <cassert>#include <set>#include <fstream>#include <iostream>#include <string>#include <sstream>#include <vector>#include <queue>#include <functional>#include <cstring>#include <algorithm>#include <cctype>#pragma comment(linker, "/STACK:102400000,102400000")#include <string>#include <map>#include <cmath>//#include <ext/pb_ds/assoc_container.hpp>//#include <ext/pb_ds/hash_policy.hpp>using namespace std;//using namespace __gnu_pbds;#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define Lowbit(x) ((x) & (-x))#define MP(a, b) make_pair(a, b)#define MS(p, num) memset(p, num, sizeof(p))#define PB push_back#define X first#define Y second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define LC rt << 1, l, mid#define RC rt << 1|1, mid + 1, r#define LRT rt << 1#define RRT rt << 1|1#define FOR(i, a, b) for (int i=(a); (i) < (b); (i)++)#define FOOR(i, a, b) for (int i = (a); (i)<=(b); (i)++)#define TRAVERSAL(u, i) for (int i = head[u]; i != -1; i = edge[i].nxt)const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const double eps = 1e-8;const int MAXN = 1e7+10;const int MOD = 1e9+7;const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };const int seed = 131;int cases = 0;typedef pair<int, int> pii;int arr[3], vis[26*26*26];queue<int> Q;string start, target;set<int> mp;void dfs(int u, int tar, int sum, int cnt){    if (sum > 26*26*26) return;    if (u == 3)    {        if (!vis[sum] && !mp.count(sum))        {            vis[sum] = cnt+1;            Q.push(sum);        }        return;    }    if (u == tar)    {        dfs(u+1, tar, sum*26 + (arr[u]+1)%26, cnt);        dfs(u+1, tar, sum*26 + (arr[u]-1+26)%26, cnt);    }    else dfs(u+1, tar, sum*26 + arr[u], cnt);}void Push(int u, int cnt){    arr[2] = u % 26; u /= 26;    arr[1] = u % 26; u /= 26;    arr[0] = u;    for (int i = 0; i <= 2; i++)        //变换第i个字母        dfs(0, i, 0, cnt);}int Hash(char a, char b, char c){    int ret = 0;    ret = a-'a'; ret *= 26;    ret += b-'a'; ret *= 26;    ret += c-'a';    return ret;}void bfs(){    int ed = Hash(target[0], target[1], target[2]);    Q.push(Hash(start[0], start[1], start[2]));    while (!Q.empty())    {        int u = Q.front(); Q.pop();        if (mp.count(u)) continue;        if (u == ed) { printf("%d\n", vis[u]); return; }        Push(u, vis[u]);    }    printf("-1\n");}int main(){    //ROP;    int T;    scanf("%d", &T);    while (T--)    {        printf("Case %d: ", ++cases);        mp.clear();        while (!Q.empty()) Q.pop();        MS(vis, 0);        cin >> start >> target;        int n;        scanf("%d", &n);        while (n--)        {            string a, b, c;            cin >> a >> b >> c;            FOR(i, 0, SZ(a)) FOR(j, 0, SZ(b)) FOR(k, 0, SZ(c)) mp.insert(Hash(a[i], b[j], c[k]));        }        bfs();    }    return 0;}