LOJ#2632. 「BalticOI 2011 Day1」打开灯泡 Switch the Lamp On

题目描述

译自 BalticOI 2011 Day1 T3「Switch the Lamp On」
有一种正方形的电路元件，在它的两组相对顶点中，有一组会用导线连接起来，另一组则不会。
有 N×M 个这样的元件，你想将其排列成 N 行 M 列放在电路板上。电路板的左上角连接电源，右下角连接灯泡。

试求：至少要旋转多少个正方形元件才能让电源与灯泡连通，若无解则输出 NO SOLUTION。

Casper is designing an electronic circuit on a N×M rectangular grid plate. There are N×M square tiles that are aligned to the grid on the plate. Two (out of four) opposite corners of each tile are connected by a wire.
A power source is connected to the top left corner of the plate. A lamp is connected to the bottom right corner of the plate. The lamp is on only if there is a path of wires connecting power source to lamp. In order to switch the lamp on, any number of tiles can be turned by 90° (in both directions).
In the picture above the lamp is off. If any one of the tiles in the second column from the right is turned by 90° , power source and lamp get connected, and the lamp is on.
Write a program to find out the minimal number of tiles that have to be turned by 90° to switch the lamp on.

输入格式

第一行有两个整数 NN和 M。
在接下来的 N 行中，每行有 M 个字符。每个字符均为 \ 或 /，表示正方形元件上导线的连接方向。

The first line of input contains two integer numbers NNN and MMM, the dimensions of the plate. In each of the following NNN lines there are MMM symbols – either \ or / – which indicate the direction of the wire connecting the opposite vertices of the corresponding tile.

输出格式

输出共一行，若有解则输出一个整数，表示至少要旋转多少个正方形元件才能让电源与灯泡连通；若无解则输出 NO SOLUTION。

There must be exactly one line of output. If it is possible to switch the lamp on, this line must contain only one integer number: the minimal number of tiles that have to be turned to switch on the lamp. If it is not possible, output the string: NO SOLUTION

样例

样例输入

3 5
\\/\\
\\///
/\\\\

样例输出

1

数据范围与提示

对于 40% 的数据，1≤N≤4,1≤M≤5。
对于所有数据，1≤N,M≤500。

题解

唔...这题$spfa+SLF$跑的飞快

因为跑bfs要判的东西貌似很多的样子所以我直接连边跑spfa了

对于能够直接走的那就连一条0边，不能直接走就连1边

然后注意坐标位置的取值...

因为边权只有0和1所以SLF在这里很有用

最后就是空间要乘个8倍

#include <bits/stdc++.h>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline 

#define in(a) a=read()
#define out(a) printf( "%d" , a )
#define outn(a) out(a),putchar('\n')

#define I_int int
inline I_int read() {

    I_int x =  , f =  ; char c = getchar() ;
    while( c < '' || c > '' ) {
        if( c == '-' ) f = - ;
        c = getchar() ;
    }
    while( c >= '' && c <= '' ) {
        x = (x << ) + (x << ) + c -  ;
        c = getchar() ;
    }
    return x * f ;
}
#undef I_int

using namespace std ;

const int N =  ;
const int M = (*+)* ;
const int dx[] = {,,,-};
const int dy[] = {-,,,};

char a[ N ][ N ] ;
int n , m ;
int head[ M ] , d[ M ] , cnt , vis[ M ] ;
struct node {
    int to , nxt , v ;
} e[ M <<  ] ;

void ins( int u , int v , int w ) {
    e[ ++ cnt ].to = v ;
    e[ cnt ].nxt = head[ u ] ;
    e[ cnt ].v = w ;
    head[ u ] = cnt ;
}

bool check( int x , int y ) {
    if( x <  || x > n || y <  || y > m ) return  ;
    return  ;
}

int zb( int x , int y ) {
    return (x-)*(m+)+y ;
}

deque<int>q;

void spfa() {
    vis[  ] =  ;
    for( int i =  ; i <= (n+)*(m+) ; i ++ ) d[ i ] = inf ;
    d[  ] =  ;
    q.push_front();
    while( !q.empty() ) {
        int u = q.front() ; q.pop_front() ;
        vis[ u ] =  ;
        for( int i = head[ u ] ; i ; i = e[ i ].nxt ) {
            int v = e[ i ].to ;
            if( d[ v ] > d[ u ] + e[ i ].v ) {
                d[ v ] = d[ u ] + e[ i ].v ;
                if( !vis[ v ] ) {
                    vis[ v ] =  ;
                    if(!e[i].v) q.push_front(v);
                    else q.push_back(v) ;
                }
            }
        }
    }
    if( d[zb(n,m)] == inf ) puts("NO SOLUTION") ;
    else outn( d[ zb(n+,m+) ] ) ;
}

int main() {
    in( n ) ; in( m ) ;
    for( int i =  ; i <= n ; i ++ ) {
        scanf( "%s" , a[ i ] +  ) ;
        for( int j =  ; j <= m ; j ++ ) {
            if( a[ i ][ j ] == '\\' ) {
                ins( zb(i,j+) , zb(i+,j) ,  ) , ins( zb(i+,j) , zb(i,j+) ,  ) ;
                ins( zb(i,j) , zb(i+,j+) ,  ) , ins( zb(i+,j+) , zb(i,j) ,  ) ;
            }
            if( a[ i ][ j ] == '/' ) {
                ins( zb(i,j+) , zb(i+,j) ,  ) , ins( zb(i+,j) , zb(i,j+) ,  ) ;
                ins( zb(i,j) , zb(i+,j+) ,  ) , ins( zb(i+,j+) , zb(i,j) ,  ) ;
            }
        }
    }
    spfa() ;
}