(A*,IDA*,DFS)eight(p1077)

经典的八数码问题，有人说不做此题人生不完整，哈哈。

状态总数是9! = 362880 种，不算太多，可以满足广搜和A*对于空间的需求。

状态可以每次都动态生成，也可以生成一次存储起来，我用的动态生成，《组合数学》书上有一种生成排列的方法叫做"序数法"，我看了一会书，把由排列到序数，和由序数到排列的两个函数写了出来，就是代码中的int order(const char *s, int n) 和void get_node(int num, node &tmp)两个函数。

启发函数，用的是除空格外的八个数字到正确位置的网格距离。

几种方法的比较：广搜，效率最低，500ms；A*，32ms，已经比较高效了；IDA*， 0ms，空间也减少许多。A*为判重付出了巨大代价，时间 and 空间，uva上还有一个15数码的题，用A*肯定会爆空间的。IDA*不记录已经走过的路径，所以省去了空间，也省去了判断重复的步骤，但是会出现重复计算。

 

广搜：

代码// BFS#include<iostream>#include<cstdio>#include<queue>using namespace std;/* 把1..n的排列映射为数字 0..(n!-1) */int fac[] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 };//...int order(const char *s, int n) {    int i, j, temp, num;    num = 0;    for (i = 0; i < n-1; i++) {        temp = 0;        for (j = i + 1; j < n; j++) {            if (s[j] < s[i])                temp++;        }        num += fac[s[i] -1] * temp;    }    return num;}bool is_equal(const char *b1, const char *b2){    for(int i=0; i<9; i++)        if(b1[i] != b2[i])            return false;    return true;}//hashstruct node{    char board[9];    char space;//空格所在位置};const int TABLE_SIZE = 362880;int hash(const char *cur){    return order(cur, 9);}/* 整数映射成排列 */void get_node(int num, node &tmp) {    int n=9;    int a[9]; //求逆序数    for (int i = 2; i <= n; ++i) {        a[i - 1] = num % i;        num = num / i;        tmp.board[i - 1] = 0;//初始化    }    tmp.board[0] = 0;    int rn, i;    for (int k = n; k >= 2; k--) {        rn = 0;        for (i = n - 1; i >= 0; --i) {            if (tmp.board[i] != 0)                continue;            if (rn == a[k - 1])                break;            ++rn;        }        tmp.board[i] = k;    }    for (i = 0; i < n; ++i)        if (tmp.board[i] == 0) {            tmp.board[i] = 1;            break;        }    tmp.space = n - a[n-1] -1;}char visited[TABLE_SIZE];int parent[TABLE_SIZE];char move[TABLE_SIZE];int step[4][2] = {{-1, 0},{1, 0}, {0, -1}, {0, 1}};//u, d, l, rvoid BFS(const node & start){    int x, y, k, a, b;    int u, v;    for(k=0; k<TABLE_SIZE; ++k)        visited[k] = 0;    u = hash(start.board);    parent[u] = -1;    visited[u] = 1;    queue<int> que;    que.push(u);    node tmp, cur;    while(!que.empty()){        u = que.front();        que.pop();        get_node(u, cur);        k = cur.space;        x = k / 3;        y = k % 3;        for(int i=0; i<4; ++i){            a = x + step[i][0];            b = y + step[i][1];            if(0<=a && a<=2 && 0<=b && b<=2){                tmp = cur;                tmp.space = a*3 + b;                swap(tmp.board[k], tmp.board[tmp.space]);                v = hash(tmp.board);                if(visited[v] != 1){                    move[v] = i;                    visited[v] = 1;                    parent[v] = u;                    if(v == 0) //目标结点hash值为0                        return;                    que.push(v);                }            }        }    }}void print_path(){    int n, u;    char path[1000];    n = 1;    path[0] = move[0];    u = parent[0];    while(parent[u] != -1){        path[n] = move[u];        ++n;        u = parent[u];    }    for(int i=n-1; i>=0; --i){        if(path[i] == 0)            printf("u");        else if(path[i] == 1)            printf("d");        else if(path[i] == 2)            printf("l");        else            printf("r");    }}int main(){    freopen("in", "r", stdin);    node start;    char c;    for(int i=0; i<9; ++i){        cin>>c;        if(c == 'x'){            start.board[i] = 9;            start.space = i;        }        else            start.board[i] = c - '0';    }    BFS(start);    if(visited[0] == 1)        print_path();    else        printf("unsolvable");    return 0;}

A*算法：

代码中对priority_queue<>模板的使用还是很有技巧性的，通过push一个最小的，再把它pop出来就解决了由于更改造成不一致性。前面这种做法是错误的，多谢一位朋友的提醒，这种方式确实不能保持堆的性质。不过我们可以采用冗余的办法，直接插入新值，这就不会破坏堆的性质了。

修改过的代码：

代码// A*#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<queue>using namespace std;/* 把1..n的排列映射为数字 0..(n!-1) */int fac[] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880 };//...int order(const char *s, int n) {    int i, j, temp, num;    num = 0;    for (i = 0; i < n-1; i++) {        temp = 0;        for (j = i + 1; j < n; j++) {            if (s[j] < s[i])                temp++;        }        num += fac[s[i] -1] * temp;    }    return num;}bool is_equal(const char *b1, const char *b2){    for(int i=0; i<9; i++)        if(b1[i] != b2[i])            return false;    return true;}//hashstruct node{    char board[9];    char space;//空格所在位置};const int TABLE_SIZE = 362880;int hash(const char *cur){    return order(cur, 9);}/* 整数映射成排列 */void get_node(int num, node &tmp) {    int n=9;    int a[9]; //求逆序数    for (int i = 2; i <= n; ++i) {        a[i - 1] = num % i;        num = num / i;        tmp.board[i - 1] = 0;//初始化    }    tmp.board[0] = 0;    int rn, i;    for (int k = n; k >= 2; k--) {        rn = 0;        for (i = n - 1; i >= 0; --i) {            if (tmp.board[i] != 0)                continue;            if (rn == a[k - 1])                break;            ++rn;        }        tmp.board[i] = k;    }    for (i = 0; i < n; ++i)        if (tmp.board[i] == 0) {            tmp.board[i] = 1;            break;        }    tmp.space = n - a[n-1] -1;}//启发函数: 除去x之外到目标的网格距离和int goal_state[9][2] = {{0,0}, {0,1}, {0,2},        {1,0}, {1,1}, {1,2}, {2,0}, {2,1}, {2,2}};int h(const char *board){    int k;    int hv = 0;    for(int i=0; i<3; ++i)        for(int j=0; j<3; ++j){            k = i*3+j;            if(board[k] != 9){                hv += abs(i - goal_state[board[k]-1][0]) +                        abs(j - goal_state[board[k] -1][1]);            }        }    return hv;}int f[TABLE_SIZE], d[TABLE_SIZE];//估计函数和深度//优先队列的比较对象struct cmp{    bool operator () (int u, int v){        return f[u] > f[v];    }};char color[TABLE_SIZE];//0, 未访问；1, 在队列中，2, closedint parent[TABLE_SIZE];char move[TABLE_SIZE];int step[4][2] = {{-1, 0},{1, 0}, {0, -1}, {0, 1}};//u, d, l, rvoid A_star(const node & start){    int x, y, k, a, b;    int u, v;    priority_queue<int, vector<int>, cmp> open;    memset(color, 0, sizeof(char) * TABLE_SIZE);    u = hash(start.board);    parent[u] = -1;    d[u] = 0;    f[u] = h(start.board);    open.push(u);    color[u] = 1;    node tmp, cur;    while(!open.empty()){        u = open.top();        if(u == 0)            return;        open.pop();        get_node(u, cur);        k = cur.space;        x = k / 3;        y = k % 3;        for(int i=0; i<4; ++i){            a = x + step[i][0];            b = y + step[i][1];            if(0<=a && a<=2 && 0<=b && b<=2){                tmp = cur;                tmp.space = a*3 + b;                swap(tmp.board[k], tmp.board[tmp.space]);                v = hash(tmp.board);                if(color[v] == 1 && (d[u] + 1) < d[v]){//v in open                    move[v] = i;                    f[v] = f[v] - d[v] + d[u] + 1;//h[v]已经求过                    d[v] = d[u] + 1;                    parent[v] = u;                    //直接插入新值, 有冗余，但不会错                    open.push(v);                }                else if(color[v] == 2 && (d[u]+1)<d[v]){//v in closed                    move[v] = i;                    f[v] = f[v] - d[v] + d[u] + 1;//h[v]已经求过                    d[v] = d[u] + 1;                    parent[v] = u;                    open.push(v);                    color[v] = 1;                }                else if(color[v] == 0){                    move[v] = i;                    d[v] = d[u] + 1;                    f[v] = d[v] + h(tmp.board);                    parent[v] = u;                    open.push(v);                    color[v] = 1;                }            }        }        color[u] = 2; //    }}void print_path(){    int n, u;    char path[1000];    n = 1;    path[0] = move[0];    u = parent[0];    while(parent[u] != -1){        path[n] = move[u];        ++n;        u = parent[u];    }    for(int i=n-1; i>=0; --i){        if(path[i] == 0)            printf("u");        else if(path[i] == 1)            printf("d");        else if(path[i] == 2)            printf("l");        else            printf("r");    }}int main(){    //freopen("in", "r", stdin);    node start;    char c;    for(int i=0; i<9; ++i){        cin>>c;        if(c == 'x'){            start.board[i] = 9;            start.space = i;        }        else            start.board[i] = c - '0';    }    A_star(start);    if(color[0] != 0)        print_path();    else        printf("unsolvable");    return 0;}

IDA*:

代码// IDA*#include<iostream>#include<cstdio>#include<cstdlib>using namespace std;#define    SIZE 3char board[SIZE][SIZE];//启发函数: 除去x之外到目标的网格距离和int goal_state[9][2] = {{0,0}, {0,1}, {0,2},        {1,0}, {1,1}, {1,2}, {2,0}, {2,1}, {2,2}};int h(char board[][SIZE]){    int cost = 0;    for(int i=0; i<SIZE; ++i)        for(int j=0; j<SIZE; ++j){            if(board[i][j] != SIZE*SIZE){                cost += abs(i - goal_state[board[i][j]-1][0]) +                        abs(j - goal_state[board[i][j]-1][1]);            }        }    return cost;}int step[4][2] = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};//u, l, r, dchar op[4] = {'u', 'l', 'r', 'd'};char solution[1000];int bound;     //上界bool ans;    //是否找到答案int DFS(int x, int y, int dv, char pre_move){// 返回next_bound    int hv = h(board);    if(hv + dv > bound)        return dv + hv;    if(hv == 0){        ans = true;        return dv;    }    int next_bound = 1e9;    for(int i=0; i<4; ++i){        if(i + pre_move == 3)//与上一步相反的移动            continue;        int nx = x + step[i][0];        int ny = y + step[i][1];        if(0<=nx && nx<SIZE && 0<=ny && ny<SIZE){            solution[dv] = i;            swap(board[x][y], board[nx][ny]);            int new_bound = DFS(nx, ny, dv+1, i);            if(ans)                return new_bound;            next_bound = min(next_bound, new_bound);            swap(board[x][y], board[nx][ny]);        }    }    return next_bound;}void IDA_star(int sx, int sy){    ans = false;    bound = h(board);//初始代价    while(!ans && bound <= 100)//上限        bound = DFS(sx, sy, 0, -10);}int main(){    freopen("in", "r", stdin);    int sx, sy;//起始位置    char c;    for(int i=0; i<SIZE; ++i)        for(int j=0; j<SIZE; ++j){            cin>>c;            if(c == 'x'){                board[i][j] = SIZE * SIZE;                sx = i;                sy = j;            }            else                board[i][j] = c - '0';        }    IDA_star(sx, sy);    if(ans){        for(int i=0; i<bound; ++i)            cout<<op[solution[i]];    }    else        cout<<"unsolvable";    return 0;}