POJ 1077 Eight

来源：互联网发布：软件测试辛苦吗编辑：程序博客网时间：2024/06/07 01:10
题目链接：http://poj.org/problem?id=1077
题意：八数码问题，求解输出一种可行的方案。
思路：很经典的bfs问题，可以用来练习双向bfs。
#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <iostream>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <sstream>#include <queue>#include <utility>using namespace std;#define rep(i,j,k) for (int i=j;i<=k;i++)#define Rrep(i,j,k) for (int i=j;i>=k;i--)#define Clean(x,y) memset(x,y,sizeof(x))#define LL long long#define ULL unsigned long long#define inf 0x7fffffff#define mod %100000007const int dx[4] = {0,0,1,-1};const int dy[4] = {1,-1,0,0};const char way2[4] = {'l','r','u','d'}; //反向方向对应的走法const char way1[4] = {'r','l','d','u'}; //正向方向对应的走法struct node{    int a[9];    int x,y;}goal,q[370000],st;map<int,int> p;      //存状态map<int,string> ans; //路径int gethash(node &x){    int ans = 0;    rep(i,0,8) ans = ans * 10 + x.a[i];    return ans;}int getpos(int x,int y){    return x*3+y;}bool check(int x,int y){    return ( x >=0 && x < 3 && y >= 0 && y < 3 );}void init() //读入和初始化{    ans.clear();    p.clear();    rep(i,0,8)    {        char c = getchar();        while( c == ' ' || c == '\n' ) c = getchar();        st.a[i] = (c=='x')?0:c-'0';        if ( st.a[i] == 0 )        {            st.x = i / 3;            st.y = i % 3;        }    }    rep(i,0,7) goal.a[i] = i + 1;    goal.a[8] = 0;    goal.x = goal.y = 2;}void bfs(){    int head,tail;    head = tail = -1;    if ( memcmp(&st , &goal , sizeof(st)) == 0 )    {        puts("");        return;    }    tail++;    memcpy( &q[tail] , &st , sizeof(st) ); //初始状态入队    tail++;    memcpy( &q[tail] , &goal , sizeof(goal) ); //目标状态入队    p[gethash(st)] = 1;  //正着拓展记为1    p[gethash(goal)] = 2;//反着拓展记为1    while( head < tail )    {        head++;        node &temp = q[head];        int xx,yy;        int temphash = gethash(temp);        rep(i,0,3)        {            xx = temp.x + dx[i];            yy = temp.y + dy[i];            if ( !check( xx , yy ) ) continue;            tail++;            node & next = q[tail];            memcpy( &next , &q[head] , sizeof(next) );            next.x = xx;            next.y = yy;            swap( next.a[ xx*3+yy ] , next.a[ temp.x*3 + temp.y ] );            //next为一个新的拓展            int hash = gethash(next);            if ( !p[hash] ) //如果next之前没有存在过            {                if ( p[temphash] == 1 )                    ans[hash] = ans[temphash] + way1[i];                else                    ans[hash] = way2[i] + ans[temphash]; //反着拓展要倒着叠加路径                p[hash] = p[temphash];  //记录路径并标记状态            }            else            {                if ( p[hash] != p[temphash] ) //如果队首拓展出来的状态和它的标记不一样，也就是正反搜索遇到了                {                    if ( p[hash] == 1 )                        cout<<ans[hash]<<way2[i]<<ans[temphash];//注意输出路径的顺序                    else                        cout<<ans[temphash]<<way1[i]<<ans[hash];                    return;                }                else tail--;//这次新的拓展是一个重复的，不保留。            }        }    }}int main(){    init();    bfs();    return 0;}
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