POJ2308 Dearboy's Puzzle(DFS+BFS)

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连连看的小游戏(只有4种牌),判断是否能全部消除。思路非常简单但是要注意减枝。
外层的DFS指定一张牌来想办法消除它,内层的BFS以指定的这张牌为起点,向四周扩展找到能和它配对的牌。然后每一对能消除的牌都考虑一遍。
1.连的线不能越界到格子外边;
2.有的牌只有奇数个,无解
3.出现了 A B 这种情况而且,AB分别只有两张,那么肯定无解;不加这个减枝会超时。
。。。。B A

#include<cstdio>#include<cstring>#include<queue>using namespace std;int mp[12][12],n,m,num[5],sum;int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};bool flag,vis[11][11];char str[12];void Init(){    memset(mp,0,sizeof mp);    memset(num,0,sizeof num);    sum = flag = 0;}struct Node{    int x,y,turn,d;    Node(){}    Node(int a,int b,int c,int dd)    {x = a; y = b; turn = c; d = dd;}};bool inarea(int x,int y){    if(x<1||y<1||x>n||y>m) return false;    else return true;}void bfs(int x,int y,int w,Node *can,int &len){    memset(vis,0,sizeof vis);    queue<Node> Q;    Q.push(Node(x,y,0,-1));    vis[x][y] = 1;    while(!Q.empty())    {        Node u = Q.front();        Q.pop();        if(mp[u.x][u.y] == w)        {            can[++len].x = u.x;            can[len].y = u.y;            continue;        }        for(int i = 0; i < 4; i++)        {            int tx = u.x+dir[i][0],ty = u.y+dir[i][1],tturn = 0;            if(!inarea(tx,ty) || (mp[tx][ty] != -1&&mp[tx][ty] != w) || vis[tx][ty]) continue;            if(u.d == i||u.d == -1)                tturn = u.turn;            else tturn = u.turn+1;            if(tturn > 2) continue;            vis[tx][ty] = 1;            Q.push(Node(tx,ty,tturn,i));        }    }}bool check(){    for(int i = 1; i < n; i++)        for(int j = 1; j < m; j++)        {            if(mp[i][j] != -1&&mp[i][j+1] != -1&&mp[i][j] != mp[i][j+1])            {                if(mp[i][j] == mp[i+1][j+1]&&mp[i][j+1] == mp[i+1][j]&&num[mp[i][j]] == 2&&num[mp[i][j+1]] == 2)                    return false;            }        }    return true;    }void dfs(int cur){    if(cur == 0)    {        flag = 1;        return;    }    if(flag) return;    if(!check()) return;//减枝    for(int i = 1; i <= n; i++)    for(int j = 1; j <= n; j++)    {        if(flag) return;        if(mp[i][j] != -1)        {            int w = mp[i][j],len = 0;            Node can[110];            mp[i][j] = -1;            bfs(i,j,w,can,len);//搜索能与(i,j)配对的牌            for(int k = 1; k <= len; k++)            {                mp[can[k].x][can[k].y] = -1;                num[w] -= 2;                dfs(cur-2);                num[w] += 2;                mp[can[k].x][can[k].y] = w;            }            mp[i][j] = w;        }    }   }int main(){    while(scanf("%d%d",&n,&m) != EOF&&n+m)    {        Init();        for(int i = 1; i <= n; i++)        {            scanf("%s",str+1);            for(int j = 1; j <= m; j++)            {                if(str[j] == '*') mp[i][j] = -1;                else mp[i][j] = str[j]-'A'+1,num[str[j]-'A'+1]++;            }        }        sum = num[1]+num[2]+num[3]+num[4];        if(num[1]%2||num[2]%2||num[3]%2||num[4]%2)//有的牌只有奇数个,无解        {               printf("no\n");            continue;        }        dfs(sum);        if(flag) printf("yes\n");        else printf("no\n");    }}
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