CodeForces1214D

CodeForces1214D
这个题据我所知有两种比较优秀的做法.
第一种是(DP)统计每个点的路径数,然后找出必经点,再从必经点开始(bfs)堵路.
第二种比较简单,你先(bfs)一遍,如果不连通,直接输出(0),否则,找到任意一条路径(可以发现,一定是最短路)堵死.
然后重复这个过程,进行了几次(bfs)就需要至少几个障碍来堵路.
这其实有点类似于网络流最大流算法(Dinic)中的每次走一条阻塞流.
正确性,从终点的两个来路考虑即可.
(Code:)

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = a ; i <= b ; ++ i)
#define per(i,a,b) for (int i = a ; i >= b ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
       while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

template < class T >
    inline void write (T x) {
       static T stk[100] , top = 0 ;
       if ( x == 0 ) { putchar ('0') ; return ; }
       if ( x < 0 ) { x = - x ; putchar ( '-' ) ; }
       while ( x ) { stk[++top] = x % 10 ; x /= 10 ; }
       while ( top ) { putchar ( stk[top--] + '0' ) ; }
       putchar ('
') ;
    }

const int N = 1e6 + 100 ;

std::queue < pii > q ;
char e[N] ;
int n , m , pre[N] ;
bool fbd[N] , mk[N] ;
int fx[] = { 0 , 1 } ;
int fy[] = { 1 , 0 } ;

inline void bfs () {
    q.push ( { 1 , 1 } ) ; mk[1] = true ;
    while ( ! q.empty () ) {
        int dx = q.front ().X , dy = q.front().Y ;
        q.pop () ; rep ( i , 0 , 1 ) {
            int tx = dx + fx[i] , ty = dy + fy[i] ;
            int cur = m  * ( tx - 1 ) + ty ;
            if ( tx < 1 || ty < 1 || tx > n || ty > m || fbd[cur] || mk[cur] || e[cur] == '#' ) continue ;
            mk[cur] = true ; pre[cur] = m * ( dx - 1 ) + dy ; q.push ( { tx , ty } ) ;
        }
    }
    return ;
}

signed main (int argc , char * argv[] ) {
    n = rint () ; m = rint () ;
    rep ( i , 1 , n ) scanf ("%s" , e + m * ( i - 1 ) + 1 ) ;
    bfs () ; if ( ! mk[m*(n-1)+m] ) { puts ("0") ; return 0 ; } MEM ( mk , 0 ) ;
    for (int i = m*(n-1)+m ; i ; i = pre[i] ) if ( i != 1 && i != m*(n-1)+m ) fbd[i] = true ;
    bfs () ; if ( ! mk[m*(n-1)+m] ) { puts ("1") ; return 0 ; } MEM ( mk , 0 ) ;
    for (int i = m*(n-1)+m ; i ; i = pre[i] ) if ( i != 1 && i != m*(n-1)+m ) fbd[i] = true ;
    bfs () ; if ( ! mk[m*(n-1)+m] ) { puts ("2") ; return 0 ; }
    return 0 ;
}