P&D 过河游戏智能帮助实现

需求

实现状态图的自动生成
讲解图数据在程序中的表示方法
利用算法实现下一步的计算

实现过程

实现状态图的自动生成&讲解图数据在程序中的表示方法

牧师与魔鬼状态可以用一个三元组（O,P,D)来表示，分别表示当前船的位置（0表示在出发地，1表示在目的地），出发地牧师数量和出发地魔鬼数量，状态数一共是2∗4∗4=322 * 4*4=322∗4∗4=32

我先用c++写了一个小程序构造出状态转移矩阵

首先求出每个状态下转移的邻接矩阵，转移表存储在数组a[32][32]a[32][32]a[32][32]中，如果状态i可以一步到达状态j，则a[i][j]=1a[i][j]=1a[i][j]=1，否则a[i][j]=0a[i][j]=0a[i][j]=0

int get(int o,int i,int j){ //将三元组转化成状态编号return o*16+i*4+j;
}
bool jud(int o,int i,int j){ //判断当前状态是否合法if (i<0||i>3||j<0||j>3) return 0;if ((i<j)&&(i!=0)) return 0;if ((3-i<3-j)&&(3-i!=0)) return 0;return 1;
}
#define rep(i,l,r) for (int i=l;i<=r;i++)
rep(o,0,1){rep(i,0,3){rep(j,0,3){int st=get(o,i,j);if (!jud(o,i,j)) continue;if (o==0){  //from 出发地的牧师和魔鬼只可能减少rep(dx,-2,0) rep(dy,-2,0) if (1<=abs(dx+dy)&&abs(dx+dy)<=2){if (!jud(o^1,i+dx,j+dy)) continue;a[get(o,i,j)][get(o^1,i+dx,j+dy)]=1;}}else { //to 出发地的牧师和魔鬼只可能增加rep(dx,0,2) rep(dy,0,2) if (1<=dx+dy&&dx+dy<=2){if (!jud(o^1,i+dx,j+dy)) continue;a[get(o,i,j)][get(o^1,i+dx,j+dy)]=1;}}}}
}

然后用最短路算法spfa求出所有状态到终点（1，0，0）的最短路距离

rep(o,0,1) rep(i,0,3) rep(j,0,3) dis[get(o,i,j)]=inf;
dis[get(1,0,0)]=0;
queue<int> q; q.push(get(1,0,0)); vis[get(1,0,0)]=1;
while (!q.empty()){int u=q.front(); q.pop(); vis[u]=0;int uo=u/16,ux=(u%16)/4,uy=u%4;rep(i,-2,2) rep(j,-2,2) {int vo=uo^1,vx=ux+i,vy=uy+j;int v=get(vo,ux+i,uy+j);if (jud(vo,vx,vy)&&a[v][u]&&dis[v]>dis[u]+1){  //如果状态v能到达状态u且能够更新状态v的距离dis[v]=dis[u]+1;if (!vis[v]) vis[v]=1,q.push(v);}}
}

我们只需要保留在最短路图上的边并输出

rep(o,0,1) rep(i,0,3) rep(j,0,3){rep(x,0,3) rep(y,0,3) {if (dis[get(o,x,y)]==dis[get(o^1,i,j)]+1&&a[get(o,x,y)][get(o^1,i,j)]) a[get(o,x,y)][get(o^1,i,j)]=1;else a[get(o,x,y)][get(o^1,i,j)]=0;}
}
rep(o,0,1) rep(i,0,3) rep(j,0,3){printf("{");rep(no,0,1) rep(x,0,3) rep(y,0,3) {printf("%d",a[get(o,i,j)][get(no,x,y)]);if (x==3&&y==3&&no==1) printf("},\n");else printf(",");}
}

状态转移矩阵

{{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}}

在游戏中通过这张状态表就可以得知下一步应该到达哪一个状态了。

利用算法实现下一步的计算

在之前作业5中（https://blog.csdn.net/ctlchild/article/details/101454426）

加入类AI，其中实现了函数tip，读取起点终点的牧师魔鬼数量，船的状态，计算出当前的状态，并根据转移表选择下一步的选择。

public int tip(CoastCon fromCoast,CoastCon toCoast,BoatCon boat) {   int fromP = 0, fromD = 0, toP = 0, toD = 0, boatst = 0;int[] fromCount = fromCoast.getCharacterNum();fromP += fromCount[0];fromD += fromCount[1];int[] toCount = toCoast.getCharacterNum ();toP += toCount[0];toD += toCount[1];int[] boatCount = boat.getCharacterNum ();if (boat.getStatus ()==-1) boatst=1;else boatst=0;if (boat.getStatus () == -1) {toP += boatCount[0]; toD += boatCount[1];   // boat at toCoast}else {fromP += boatCount[0]; fromD += boatCount[1];  // boat at fromCoast    }Debug.Log(fromP.ToString()+' '+fromD.ToString()+' '+get(boatst,fromP,fromD).ToString());for (int o=0;o<=1;o++)for (int i=0;i<=3;i++) for (int j=0;j<=3;j++) if (trans[get(boatst,fromP,fromD),get(o,i,j)]==1) return get(boatst,i,j);return -1;
}

运行效果

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