poj1077八数码问题——境界六（IDA*）

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迭代加深
#include<stdio.h>#include<string.h>#include<string>#include<algorithm>#include<iostream>#include<set>#include<map>#include<queue>#include<stack>#include<vector>#include<math.h>#include<stdlib.h>using namespace std;#define M 400000struct node {char s[20];int pos;}st;int h[4]={-1,0,1,0};int z[4]={0,1,0,-1};char fa[]={"urdl"};int path[M],limit;int geth(char *s)//取得h，为当前位置与目标位置的manhattan距离 {int i,j,h=0,x,y,x1,y1;for(i=0;i<9;i++){if(s[i]!='0'){x=i/3;y=i%3;x1=(s[i]-'0'-1)/3;y1=(s[i]-'0'-1)%3;h+=abs(x-x1)+abs(y-y1);}}return h;}bool isAns(node t){int i,j;if(strcmp(t.s,"123456780")==0)return 1;return 0;}int judge(int x,int y){    return (x>=0&&x<3&&y>=0&&y<3);}int IDAstar(node t,int len){if(len==limit){return isAns(t);}int i,j,k,x,y;for(i=0;i<4;i++){if(len>0&&abs(i-path[len-1])==2)continue;node tp=t;x=tp.pos/3;y=tp.pos%3;x+=h[i];y+=z[i];if(judge(x,y)){tp.pos=x*3+y;swap(tp.s[tp.pos],tp.s[t.pos]);path[len]=i;if(geth(tp.s)+len<=limit&&IDAstar(tp,len+1))return 1;}}return 0;}void output(){int i,j,k;for(i=0;i<limit;i++)printf("%c",fa[path[i]]);puts("");}int main(){int i,j,k;     char c;  for(i=0;i<9; )//读取字符串 {         scanf("%c",&c);    if(c == 'x')    {             st.s[i]='0';           st.pos=i;           i++;    }            else if(c>='0'&&c<='9')        {        st.s[i] = c;         i++;    }           else    continue;              }st.s[9]='\0';limit=geth(st.s);while(!IDAstar(st,0))limit++;output();return 0;}