UVA 1601 双向BFS

但是我们还不是很清楚每一次的状态怎么储存？我们可以用一个结构体，将每次的位置存起来，但是这个程序中用了一个更好的储存方法：我们知道最大的格数是16*16个，也就是256个，那么我们转换为二进制表示就是8位数，那么我们可以使用24位的二进制表示啊！然后我们再进行解压缩，所以这就是很神奇的地方！

普通BFS

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#define INF 0x3f3f3f3f
//const int maxn = 1e6 + 5;
const double PI = acos(-1.0);
typedef long long ll;
using namespace std;

const int maxn = 150;
const int maxs = 20;
const int dx[] = { 1,-1,0,0,0 };
const int dy[] = { 0,0,1,-1,0 };

inline int ID(int a, int b, int c) {
    return (a << 16) | (b << 8) | c;
}

int s[3], t[3];

int deg[maxn];    //记录每个编号为i的空格周围可以走的步数
int G[maxn][5];   

inline bool conflict(int a, int b, int a2, int b2) {
    return    a2 == b2 || (a2 == b && b2 == a);
}

int d[maxn][maxn][maxn];

int bfs() {
    queue<int> q;
    memset(d, -1, sizeof d);
    q.push(ID(s[0], s[1], s[2]));
    d[s[0]][s[1]][s[2]] = 0;
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        int a = (u >> 16) & 0xff, b = (u >> 8) & 0xff, c = u & 0xff;   //解码出三个鬼的位置
        if (a == t[0] && b == t[1] && c == t[2]) return d[a][b][c];
        for (int i = 0; i < deg[a]; i++) {
            int a2 = G[a][i];
            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 < deg[c]; k++) {
                    int c2 = G[c][k];
                    if (conflict(a, c, a2, c2)) continue;
                    if (conflict(b, c, b2, c2)) continue;
                    if (d[a2][b2][c2] != -1) continue;
                    d[a2][b2][c2] = d[a][b][c] + 1;
                    q.push(ID(a2, b2, c2));
                }
            }
        }
    }
    return -1;
}

int main() {
    int w, h, n;

    while (scanf("%d%d%d", &w, &h, &n) == 3 && n) {
        char maze[20][20];
        for (int i = 0; i < h; i++) fgets(maze[i], 20, stdin);

        int cnt, x[maxn], y[maxn], id[maxs][maxs];
        cnt = 0;
        for (int i = 0; i < h; i++) {
            for(int j=0;j<w;j++)
                if (maze[i][j] != ‘#‘) {
                    x[cnt] = i;
                    y[cnt] = j;
                    id[i][j] = cnt;
                    if (islower(maze[i][j])) s[maze[i][j] - ‘a‘] = cnt;
                    else if (isupper(maze[i][j])) t[maze[i][j] - ‘A‘] = cnt;
                    cnt++;
                }
        }

        for (int i = 0; i < cnt; i++) {
            deg[i] = 0;
            for (int dir = 0; dir < 5; dir++) {
                int xx = x[i] + dx[dir], yy = y[i] + dy[dir];
                if (maze[xx][yy] != ‘#‘) G[i][deg[i]++] = id[xx][yy];
            }
        }

        if (n <= 2) {
            deg[cnt] = 1;
            G[cnt][0] = cnt;
            s[2] = t[2] = cnt++;
        }
        if (n <= 1) {
            deg[cnt] = 1;
            G[cnt][0] = cnt;
            s[1] = t[1] = cnt++;
        }

        printf("%d\n", bfs());

    }
    return 0;
}

双向BFS

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#define INF 0x3f3f3f3f
//const int maxn = 1e6 + 5;
const double PI = acos(-1.0);
typedef long long ll;
using namespace std;

const int maxn = 150;
const int maxs = 20;
const int dx[] = { 1,-1,0,0,0 };
const int dy[] = { 0,0,1,-1,0 };

inline int ID(int a, int b, int c) {
    return (a << 16) | (b << 8) | c;
}

int s[3], t[3];

int deg[maxn];    //记录每个编号为i的空格周围可以走的步数
int G[maxn][5];
char maze[maxn][maxn];
int color[maxn][maxn][maxn];

inline bool conflict(int a, int b, int a2, int b2) {
    //两个鬼是exchange位置（违反第2条）
    //两个鬼移动到同一个格子（违反第1条）
    return    a2 == b2 || (a2 == b && b2 == a);
}

int d1[maxn][maxn][maxn];

int bfs() {
    queue<int> qf;
    queue<int> qb;

    d1[s[0]][s[1]][s[2]] = 0;
    d1[t[0]][t[1]][t[2]] = 1;

    qf.push(ID(s[0], s[1], s[2]));
    qb.push(ID(t[0], t[1], t[2]));

    while (!qf.empty() || !qb.empty()) {
        int fnum = qf.size(), bnum = qb.size();
        while (fnum--) {
            int u = qf.front(); qf.pop();
            int a = (u >> 16) & 0xff, b = (u >> 8) & 0xff, c = u & 0xff;

            for (int i = 0; i < deg[a]; i++) {
                int a2 = G[a][i];
                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 < deg[c]; k++) {
                        int c2 = G[c][k];
                        if (conflict(a, c, a2, c2) || conflict(b, c, b2, c2)) continue;
                        if (color[a2][b2][c2] == 0) {
                            d1[a2][b2][c2] = d1[a][b][c] + 1;
                            color[a2][b2][c2] = 1;
                            qf.push(ID(a2, b2, c2));
                        }
                        else if (color[a2][b2][c2] == 2) {
                            return d1[a][b][c] + d1[a2][b2][c2];
                        }
                    }
                }
            }
        }
        while (bnum--) {
            int u = qb.front(); qb.pop();
            int a = (u >> 16) & 0xff, b = (u >> 8) & 0xff, c = u & 0xff;

            for (int i = 0; i < deg[a]; i++) {
                int a2 = G[a][i];
                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 < deg[c]; k++) {
                        int c2 = G[c][k];
                        if (conflict(a, c, a2, c2) || conflict(b, c, b2, c2)) continue;
                        if (color[a2][b2][c2] == 0) {
                            d1[a2][b2][c2] = d1[a][b][c] + 1;
                            color[a2][b2][c2] = 2;
                            qb.push(ID(a2, b2, c2));
                        }
                        else if (color[a2][b2][c2] == 1) {
                            return d1[a][b][c] + d1[a2][b2][c2];
                        }
                    }
                }
            }
        }
    }
    return -1;
}

int main() {
    int w, h, n;
    while (scanf("%d%d%d", &w, &h, &n) == 3, n) {
        for (int i = 0; i < h; i++) fgets(maze[i], 20, stdin);
        int cnt = 0;
        int x[maxn], y[maxn];
        int id[maxs][maxs];
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                if (maze[i][j] != ‘#‘) {
                    x[cnt] = i, y[cnt] = j, id[i][j] = cnt;
                    if (islower(maze[i][j])) s[maze[i][j] - ‘a‘] = cnt;
                    else if (isupper(maze[i][j])) t[maze[i][j] - ‘A‘] = cnt;
                    cnt++;
                }
            }
        }

        for (int i = 0; i < cnt; i++) {
            for (int j = 0; j < 5; j++) {
                int xx = x[i] + dx[j], yy = y[i] + dy[j];
                if (maze[xx][yy] != ‘#‘) G[i][deg[i]++] = id[xx][yy];
            }
        }

        if (n <= 2) {
            deg[cnt] = 1;
            G[cnt][0] = cnt;
            s[2] = t[2] = cnt++;
        }
        if (n <= 1) {
            deg[cnt] = 1;
            G[cnt][0] = cnt;
            s[1] = t[1] = cnt++;
        }

        memset(d1, 0, sizeof d1);
        memset(color, 0, sizeof color);

        if (s[0] == t[0] && s[1] == t[1] && s[2] == t[2]) printf("0\n");
        else printf("%d\n", bfs());
    }
    return 0;
}

原文地址：https://www.cnblogs.com/hznumqf/p/12358464.html