hdu 4490 Mad Veterinarian（bfs）

有三種物品，每個物品可以變換成一個或多個其他物品。這種變換關係是可逆的。

然後求從初始狀態轉換到目標狀態所需的最少步數。直接bfs就能搞了，記錄狀態的話小hash一發話就行了。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<fstream>
#include<sstream>
#include<cstdlib>
#include<vector>
#include<string>
#include<cstdio>
#include<bitset>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<set>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>=b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define debug puts("**debug**")
#define LL long long
#define PB push_back
#define MP make_pair
using namespace std;

const int maxm = 1e5;
char str[10] = "ABCabc";
/***************012345************/
int on[maxm], trans[5][5], pre[2000000], path[2000000];
bool vis[200][200][200];

struct Node
{
    int v[3], steps;
    void get()
    {
        REP(i, 3) scanf("%d", &v[i]);
        steps = 0;
    }
    bool operator < (const Node& rhs) const
    {
        return steps > rhs.steps;
    }
}S, T;

inline int hash(Node x)
{
    return x.v[0]*10008+x.v[1]*108+x.v[2];
}

bool move(Node x, Node& y, int type)
{
    y = x;
    if(type < 3)
    {
        if(x.v[type] > 0)
        {
            y.v[type]--;
            REP(i, 3) y.v[i] += trans[type][i];
            return !vis[y.v[0]][y.v[1]][y.v[2]];
        }
        return false;
    }
    else
    {
        type -= 3;
        REP(i, 3) if(y.v[i] < trans[type][i]) return false;
        REP(i, 3) y.v[i] -= trans[type][i];
        y.v[type]++;
        return !vis[y.v[0]][y.v[1]][y.v[2]];
    }
    return false;
}

bool bfs()
{
    CLR(vis, 0); vis[S.v[0]][S.v[1]][S.v[2]] = 1;
    CLR(pre, -1);
    priority_queue<Node> q; q.push(S);
    while(!q.empty())
    {
        Node x = q.top(), y; q.pop();
        if(x.v[0] == T.v[0] && x.v[1] == T.v[1] && x.v[2] == T.v[2])
        {
            T.steps = x.steps;
            return true;
        }
        REP(i, 3) if(x.v[i] > 108) return false;
        REP(i, 6) if(move(x, y, i))
        {
            y.steps = x.steps + 1;
            pre[hash(y)] = hash(x);
            path[hash(y)] = i;
            vis[y.v[0]][y.v[1]][y.v[2]] = 1;
            q.push(y);
        }
    }
    return false;
}

void print(int x)
{
    if(pre[x] == -1) return ;
    print(pre[x]);
    putchar(str[path[x]]);
}

int P, id, cas, q;
int main()
{
   // freopen("input.txt", "r", stdin);
   // freopen("output.txt", "w", stdout);
    scanf("%d", &P);
    while(P--)
    {
        scanf("%d%d", &cas, &q);
        REP(i, 3) REP(j, 3) scanf("%d", &trans[i][j]);
        printf("%d %d\n", cas, q);
        while(q--)
        {
            scanf("%d ", &id);
            printf("%d ", id);
            S.get(); T.get();
            if(!bfs()) puts("NO SOLUTION");
            else
            {
                printf("%d ", T.steps);
                print(hash(T));
                puts("");
            }
        }
    }
    return 0;
}