codevs1049

题目地址：http://codevs.cn/problem/1049/

分析：

只要判断其是否连通就行了，找到连通的最少步数。

代码：

#include <stdio.h>
#include <string.h>
#include <set>
#include <iostream>
using namespace std;
typedef int State[5][5];
State map,sta[500005],visit;
int sub[25][2];
int step[500005],nextx[4] = { 1, 0, -1, 0 }, nexty[4] = { 0, 1, 0, -1 };
set<string> vis;
int check ( State t )
{
    char str[50];
    int count = 0;
    for ( int i = 0; i < 5; i ++ )
        for ( int j = 0; j < 5; j ++ )
            str[count++] = ( t[i][j]+'0' );
    str[count] = '\0';
    if ( vis.count ( str ) )
        return 0;
    vis.insert ( str );
    return 1;
}
void dfs ( State t, int x, int y )
{
    if ( x < 0 || x >= 5 || y < 0 || y >= 5 || visit[x][y] == 1 || t[x][y] == 0 )
        return ;
    visit[x][y] = 1;
    for ( int i = 0; i < 4; i ++ )
        dfs ( t, x+nextx[i], y+nexty[i] );
}
void print ( State t )
{
    int i,j;
    for ( i = 0; i < 5; i ++ )
    {
        for ( j = 0; j < 5; j ++ )
            printf ( "%d ", t[i][j] );
        printf ( "\n" );
    }
    printf ( "\n" );
}
int is_ok ( State t )
{
    int i;
    memset ( visit, 0, sizeof ( visit ) );
    for ( i = 0; i < 25; i ++ )
        if ( t[i/5][i%5] )
            break ;
    dfs ( t, i/5, i%5 );
    for ( i = 0; i < 25; i ++ )
        if ( visit[i/5][i%5] == 0 && t[i/5][i%5] == 1 )
            return 0;
    return 1;
}
int bfs ( )
{
    int front,rear,i,j,count;
    vis.clear ( );
    front = 0;
    rear = 1;
    memcpy ( sta[front], map, sizeof ( map ) );
    step[front] = 0;
    while ( front < rear )
    {
        count = 0;
        if ( is_ok ( sta[front] ) )
            return step[front];
        for ( i = 0; i < 5; i ++ )
            for ( j = 0; j < 5; j ++ )
                if ( sta[front][i][j] )
                {
                    sub[count][0] = i;
                    sub[count][1] = j;
                    count ++;
                }
        for ( i = 0; i < count; i ++ )
            for ( j = 0; j < 4; j ++ )
            {
                int nx = sub[i][0]+nextx[j];
                int ny = sub[i][1]+nexty[j];
                if ( nx < 0 || nx >= 5 || ny < 0 || ny >= 5 || sta[front][nx][ny] )
                    continue ;
                memcpy ( sta[rear], sta[front], sizeof ( sta[front] ) );
                sta[rear][nx][ny] = 1;
                if ( check ( sta[rear] ) )
                {
                    step[rear] = step[front]+1;
                    rear ++;
                }
            }
        front ++;
    }
    return -1;
}
int main ( )
{
    int i,j;
    char str[10];
    for ( i = 0; i < 5; i ++ )
    {
        scanf ( "%s", str );
        for ( j = 0; j < 5; j ++ )
            map[i][j] = str[j]-'0';
    }
    printf ( "%d", bfs ( ) );
    return 0;
}