POJ1077 Eight —— IDA*算法

来源：互联网发布：淘宝客服售后绩效模板编辑：程序博客网时间：2024/05/19 03:23

主页面：http://blog.csdn.net/dolfamingo/article/details/77825569

代码一：像BFS那样，把棋盘数组放在结构体中。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <map>#include <string>#include <set>#define ms(a,b) memset((a),(b),sizeof((a)))using namespace std;typedef long long LL;const int INF = 2e9;const LL LNF = 9e18;const int MOD = 1e9+7;const int MAXN = 1e6+10;#define AIM  1     //123456789的哈希值为1struct node{    int s[9];    int loc;};int fac[9] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320};int dir[4][2] = { -1,0, 1,0, 0,-1, 0,1 };char op[4] = {'u', 'd', 'l', 'r'  };int cantor(int s[])  //获得哈希函数值{    int sum = 0;    for(int i = 0; i<9; i++)    {        int num = 0;        for(int j = i+1; j<9; j++)            if(s[j]<s[i]) num++;        sum += num*fac[8-i];    }    return sum+1;}int dis_h(int s[])  //获得曼哈顿距离{    int dis = 0;    for(int i = 0; i<9; i++)    if(s[i]!=9)    {        int x = i/3, y = i%3;        int xx = (s[i]-1)/3, yy = (s[i]-1)%3;        dis += abs(x-xx) + abs(y-yy);    }    return dis;}char path[100];bool IDAstar(node now, int depth, int pre, int limit){    if(dis_h(now.s)==0)     //搜到123456789    {        path[depth] = '\0';        puts(path);        return true;    }    int x = now.loc/3;    int y = now.loc%3;    for(int i = 0; i<4; i++)    {        if(i+pre==1 || i+pre==5) continue;  //方向与上一步相反， 剪枝        int xx = x + dir[i][0];        int yy = y + dir[i][1];        if(xx>=0 && xx<=2 && yy>=0 && yy<=2)        {            node tmp = now;            tmp.s[x*3+y] = tmp.s[xx*3+yy];            tmp.s[xx*3+yy] = 9;            tmp.loc = xx*3+yy;            path[depth] = op[i];            if(depth+1+dis_h(tmp.s)<=limit)     //在限制内                if(IDAstar(tmp, depth+1, i, limit))                    return true;        }    }    return false;}int main(){    char str[50];    while(gets(str))    {        node now;        int cnt = 0;        for(int i = 0; str[i]; i++)        {            if(str[i]==' ') continue;            if(str[i]=='x') now.s[cnt] = 9, now.loc = cnt++;            else  now.s[cnt++] = str[i]-'0';        }        int limit;        for(limit = 0; limit<=100; limit++)     //迭代加深搜            if(IDAstar(now, 0, INF, limit))                break;        if(limit>100)            puts("unsolvable");    }}

代码二：根据DFS的特点，由于棋盘每次只交换一对，所以可以只开一个棋盘数组。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <map>#include <string>#include <set>#define ms(a,b) memset((a),(b),sizeof((a)))using namespace std;typedef long long LL;const int INF = 2e9;const LL LNF = 9e18;const int MOD = 1e9+7;const int MAXN = 1e6+10;int M[100];int fac[9] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320};int dir[4][2] = { -1,0, 1,0, 0,-1, 0,1 };char op[4] = {'u', 'd', 'l', 'r'  };int cantor(int s[])  //获得哈希函数值{    int sum = 0;    for(int i = 0; i<9; i++)    {        int num = 0;        for(int j = i+1; j<9; j++)            if(s[j]<s[i]) num++;        sum += num*fac[8-i];    }    return sum+1;}int dis_h(int s[])  //获得曼哈顿距离{    int dis = 0;    for(int i = 0; i<9; i++)    if(s[i]!=9)    {        int x = i/3, y = i%3;        int xx = (s[i]-1)/3, yy = (s[i]-1)%3;        dis += abs(x-xx) + abs(y-yy);    }    return dis;}char path[100];bool IDAstar(int loc, int depth, int pre, int limit){    int h = dis_h(M);    if(depth+h>limit)        return false;    if(h==0)     //搜到123456789    {        path[depth] = '\0';        puts(path);        return true;    }    int x = loc/3;    int y = loc%3;    for(int i = 0; i<4; i++)    {        if(i+pre==1 || i+pre==5) continue;  //方向与上一步相反， 剪枝        int xx = x + dir[i][0];        int yy = y + dir[i][1];        if(xx>=0 && xx<=2 && yy>=0 && yy<=2)        {            swap(M[loc], M[xx*3+yy]);            path[depth] = op[i];            if(IDAstar(xx*3+yy, depth+1, i, limit))                return true;            swap(M[loc], M[xx*3+yy]);        }    }    return false;}int main(){    char str[50];    while(gets(str))    {        int cnt = 0, loc;        for(int i = 0; str[i]; i++)        {            if(str[i]==' ') continue;            if(str[i]=='x') M[cnt] = 9, loc = cnt++;            else  M[cnt++] = str[i]-'0';        }        int limit;        for(limit = 0; limit<=100; limit++)     //迭代加深搜            if(IDAstar(loc, 0, INF, limit))                break;        if(limit>100)            puts("unsolvable");    }}

阅读全文

0 0