UVA 810 A Dicey Promblem 筛子难题 （暴力BFS+状态处理）

读懂题意以后还很容易做的，

和AbbottsRevenge类似加一个维度，筛子的形态，可以用上方的点数u和前面的点数f来表示，相对的面点数之和为7，可以预先存储u和f的对应右边的点数，点数转化就很容易了。

具体做法看代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 11;
const int maxd = 7;
int g[maxn][maxn];
bool vis[maxn][maxn][maxd][maxd];

struct State
{
    int x,y,u,f,pre;
}Q[6000];

int tor[maxd][maxd];

inline void trans(int u,int f,int i,int &nu,int &nf)
{
    if(i == 0){ nu = f; nf = 7-u; return; }
    if(i == 1){ nu = 7 - tor[u][f]; nf = f; return; }
    if(i == 2){ nu = 7 - f; nf = u ; return; }
    nu = tor[u][f]; nf = f;
}

vector<int> ans;
void print_ans(int rst){
    for(int i = rst ; ~i; i = Q[i].pre){
        ans.push_back(i);
    }
    vector<int>::reverse_iterator ri = ans.rbegin(),ed = ans.rend();
    int cnt = 1;
    for( ed--; ri != ed; ri++,cnt++){
        printf("(%d,%d),",Q[*ri].x,Q[*ri].y);
        if(!(cnt%9)) printf("
  ");
    }
    printf("(%d,%d)
",Q[*ed].x,Q[*ed].y);
}

void bfs(int sx,int sy,int su,int sf)
{
    const int dx[] = {-1,0,1,0};
    const int dy[] = {0,1,0,-1};
    int head,rear;
    head = rear = 0;
    Q[rear].u = su; Q[rear].f = sf; Q[rear].x = sx; Q[rear].y = sy; Q[rear].pre = -1;
    rear++;
    while(head<rear){
        State &u = Q[head];
        for(int i = 0; i < 4; i++){
            int nx = u.x + dx[i], ny = u.y + dy[i];
            if(~g[nx][ny] && u.u != g[nx][ny]) continue;//0也不满足条件
            if(nx == sx && ny == sy){
                Q[rear].x = nx; Q[rear].y = ny;
                Q[rear].pre = head;
                ans.clear();
                print_ans(rear);
                return;
            }
            int nu,nf;
            trans(u.u,u.f,i,nu,nf);
            if(vis[nx][ny][nu][nf]) continue;
            vis[nx][ny][nu][nf] = true;
            Q[rear].u = nu; Q[rear].f = nf; Q[rear].x = nx; Q[rear].y = ny; Q[rear].pre = head;
            rear++;
        }
        head++;
    }
    printf("No Solution Possible
");
}

int main()
{
   // freopen("in.txt","r",stdin);
    char str[25];
    tor[1][2] = 3; tor[1][3] = 5; tor[1][5] = 4; tor[1][4] = 2;
    tor[2][1] = 4; tor[2][3] = 1; tor[2][4] = 6; tor[2][6] = 3;
    tor[3][6] = 5; tor[3][5] = 1; tor[3][1] = 2; tor[3][2] = 6;
    tor[4][6] = 2; tor[4][2] = 1; tor[4][1] = 5; tor[4][5] = 6;
    tor[5][6] = 4; tor[5][4] = 1; tor[5][1] = 3; tor[5][3] = 6;
    tor[6][4] = 5; tor[6][5] = 3; tor[6][3] = 2; tor[6][2] = 4;
    const int bsz = sizeof(bool)*maxn*49;
    const int isz = sizeof(int)*maxn;
    while(~scanf("%s",str)&& strcmp(str,"END")){
        printf("%s
  ",str);
        int R,C,sx,sy,su,sf;
        scanf("%d%d%d%d%d%d",&R,&C,&sx,&sy,&su,&sf);
        memset(vis,0,bsz*(R+1));
        for(int i = 1; i <= R; i ++){
            for(int j = 1; j <= C; j++)
                scanf("%d",g[i]+j);
        }
        memset(g[R+1],0,isz);
        for(int i = 1; i <= R; i++) g[i][C+1] = 0;
        bfs(sx,sy,su,sf);
    }
    return 0;
}