The Morning after Halloween（POJ

给出w*h的网格，相当于迷宫，有大写字母和小写字母，算出小写字母走到大写字母状态时的最少步数。

另有双向搜索解和A*解

#include<cstdio>#include<queue>#include<cstring>//#include<vector>#include<cctype>using namespace std;const int maxs = 20, maxn = 150;//char map[maxs][maxs];    //原图int d[maxn][maxn][maxn];  //隐式图状态结点属性int s[4], t[4];  //记录起点与终点对应隐式图状态编号const int dx[] = { 1,-1,0,0,0 }; // 4 moves, plus "no move"const int dy[] = { 0,0,1,-1,0 };int id[maxs][maxs]; //给新图中结点编号，同时储存其坐标作为属性//用数组存储而非结构体以达到双向查询目的int x[maxn], y[maxn];//vector<int> G[maxn];     //从原图中提取可行路径作为新图int deg[maxn], G[maxn][5]; //用vector超时，此处用deg储存size，G储存上下左右中可行结点序号inline int ID(int a, int b, int c) //将状态隐式图结点编号化为一个整数{    return (a << 16) | (b << 8) | c;}inline bool conflict(int a, int b, int a2, int b2) {    return a2 == b2 || (a2 == b && b2 == a);}int bfs(){    queue<int> q;    memset(d, -1, sizeof(d)); //将结点属性初始化为-1，用于判重    d[s[0]][s[1]][s[2]] = 0;    q.push(ID(s[0],s[1],s[2]));    while (!q.empty())    {        int u = q.front();        q.pop();        int a = (u >> 16) & 0xff, b = (u >> 8) & 0xff, c = u & 0xff; //提取隐式图结点编号        if (a == t[0] && b == t[1] && c == t[2])             return d[a][b][c]; // 终点退出bfs函数        //for (int i = 0; i < G[a].size(); i++)        for (int i = 0; i < deg[a]; i++)//隐式图结点移动遍历        {            int a2 = G[a][i];            //for (int j = 0; j< G[b].size(); j++)            for (int j = 0; j < deg[b]; j++)            {                int b2 = G[b][j];                if (conflict(a, b, a2, b2)) continue;                //for (int k = 0; k < G[c].size(); k++)                for (int k = 0; k < deg[c]; k++)                {                    int c2 = G[c][k];                    if (conflict(a, c, a2, c2)) continue;                    if (conflict(c, b, c2, b2)) continue;                    if (d[a2][b2][c2] != -1) continue; //判重                    d[a2][b2][c2] = d[a][b][c] + 1; //遍历隐式图结点属性变化                    q.push(ID(a2, b2, c2));                }            }        }    }    return -1;}int main(){    int w, h, n;    while (scanf("%d%d%d\n", &w, &h, &n) && w)//用到fgets则需读取\n    {        memset(map,0,sizeof(map));        memset(x, 0, sizeof(x));        memset(y, 0, sizeof(y));        memset(id, 0, sizeof(id));        memset(deg, 0, sizeof(deg));        //for (int i = 0; i < maxn; i++)            //G[i].clear();        for (int i = 0; i < h; i++)            fgets(map[i], 20, stdin);        int cnt = 0;//用于给新图编号        for(int i=0;i<h;i++)//读入处理新图            for (int j = 0; j < w; j++)             {                if (map[i][j] != '#')                {                    if (islower(map[i][j]))                    {                        s[map[i][j] - 'a'] = cnt;                    }                    else if (isupper(map[i][j]))                    {                        t[map[i][j] - 'A'] = cnt;                    }                    x[cnt] = i;                    y[cnt] = j;                    id[i][j] = cnt++;//from 0 //给新图结点编号，记录属性（坐标）                }            }        for (int i = 0; i < cnt; i++)//建新图        {            for (int dir = 0; dir < 5; dir++)            {                int nx = x[i] + dx[dir];                int ny = y[i] + dy[dir];                if (map[nx][ny] != '#')                {                    //G[i].push_back(id[nx][ny]);                    G[i][deg[i]++] = id[nx][ny];                }            }        }        if (n <= 2) //当输入情况小于三时，创造虚拟结点，使得针对于三的程序依然可行        {             deg[cnt] = 1;             G[cnt][0] = cnt;            //G[cnt].push_back(cnt);            s[2] = t[2] = cnt++;         }        if (n <= 1)         {             deg[cnt] = 1;             G[cnt][0] = cnt;            //G[cnt].push_back(cnt);            s[1] = t[1] = cnt++;         }        printf("%d\n", bfs());    }    return 0;}