[A*] HDU 1043

来源：互联网发布：电脑连上网却没有网络编辑：程序博客网时间：2024/06/06 02:51
虽然是过了
但因为一开始hash值存在结构体里所以TLE
下次还是打表吧
#include <algorithm>#include <iostream>#include <math.h>#include <queue>#include <stack>#include <stdio.h>#include <string.h>using namespace std;struct Node {        char map[ 3 ][ 3 ];        int x, y;        int g, h; // g是已经走过的步数，h是剩下距离终点的距离        bool operator< ( const Node t ) const { return h + g > t.h + t.g; }} cur, nex;char str[ 100 ];int check_rev () //逆序数对，判断不可能的状况{        int i, j, k;        int s[ 20 ];        int cnt = 0;        for ( i = 0; i < 3; i++ ) {                for ( j = 0; j < 3; j++ ) {                        s[ 3 * i + j ] = cur.map[ i ][ j ];                        if ( s[ 3 * i + j ] == 'x' )                                continue;                        for ( k = 3 * i + j - 1; k >= 0; k-- ) {                                if ( s[ k ] == 'x' )                                        continue;                                if ( s[ k ] > s[ 3 * i + j ] )                                        cnt++;                        }                }        }        if ( cnt % 2 )                return 0;        return 1;}//康托展开 0~8的阶乘int factorial[ 9 ] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};int get_hash ( Node a ) //康托{        int i, j, k;        int s[ 20 ];        int ans = 0;        for ( i = 0; i < 3; i++ ) {                for ( j = 0; j < 3; j++ ) {                        s[ 3 * i + j ] = a.map[ i ][ j ];                        int cnt = 0;                        for ( k = 3 * i + j - 1; k >= 0; k-- ) {                                if ( s[ k ] > s[ 3 * i + j ] )                                        cnt++;                        }                        ans = ans + factorial[ i * 3 + j ] * cnt;                }        }        return ans;}// 3*3每个点的indxint pos[][ 2 ] = {{0, 0}, {0, 1}, {0, 2}, {1, 0}, {1, 1}, {1, 2}, {2, 0}, {2, 1}, {2, 2}};int get_h ( Node a ) //得到距离终点距离的值{        int i, j;        int ans = 0;        for ( i = 0; i < 3; i++ ) {                for ( j = 0; j < 3; j++ ) {                        if ( a.map[ i ][ j ] == 'x' )                                continue;                        int k = a.map[ i ][ j ] - '1';                        ans += abs ( pos[ k ][ 0 ] - i ) + abs ( pos[ k ][ 1 ] - j );                }        }        return ans;}//判断是否出界inline bool path ( int x, int y ) {        if ( x >= 0 && x < 3 && y >= 0 && y < 3 )                return true;        return false;}bool vis[ 500000 ];int dir[ 4 ][ 2 ] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};//回溯用char ch[ 500000 ];int pre[ 500000 ];char way[ 10 ] = "drul";void A_star () {        memset ( vis, 0, sizeof ( vis ) );        cur.g = 0;        cur.h = get_h ( cur );        priority_queue<Node> Q;        Q.push ( cur );        while ( !Q.empty () ) {                Node a = Q.top ();                Q.pop ();                int k_s = get_hash ( a );                for ( int i = 0; i < 4; i++ ) {                        nex = a;                        nex.x += dir[ i ][ 0 ];                        nex.y += dir[ i ][ 1 ];                        if ( path ( nex.x, nex.y ) ) {                                // swap                                nex.map[ a.x ][ a.y ] = a.map[ nex.x ][ nex.y ];                                nex.map[ nex.x ][ nex.y ] = 'x';                                nex.g += 1;                                nex.h = get_h ( nex );                                //找个变量存一下( 存在结构体里会TLE )                                int k_n = get_hash ( nex );                                if ( !vis[ k_n ] ) {                                        vis[ k_n ] = true;                                        Q.push ( nex );                                        pre[ k_n ] = k_s;                                        ch[ k_n ] = way[ i ];                                        if ( k_n == 0 )                                                return;                                }                        }                }        }}int main () {        int i, len, x, y;        //输入        while ( cin.getline ( str, 100 ) ) {                x = y = 0;                len = strlen ( str );                for ( i = 0; i < len; i++ ) {                        if ( ( str[ i ] >= '0' && str[ i ] <= '9' ) || str[ i ] == 'x' ) {                                cur.map[ x ][ y ] = str[ i ];                                if ( cur.map[ x ][ y ] == 'x' ) {                                        cur.x = x;                                        cur.y = y;                                }                                y++;                                if ( y == 3 ) {                                        y = 0;                                        x++;                                }                        }                }                //判断逆序对数，能否转换成最后结果                if ( !check_rev () ) {                        printf ( "unsolvable\n" );                        continue;                }                int sa = get_hash ( cur );                A_star ();                //输出                stack<char> s;                int now = 0;                while ( sa != now ) {                        s.push ( ch[ now ] );                        now = pre[ now ];                }                while ( !s.empty () ) {                        putchar ( s.top () );                        s.pop ();                }                printf ( "\n" );        }        return 0;}
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