zoukankan      html  css  js  c++  java
  • HDU 4285 circuits( 插头dp , k回路 )

    circuits

    Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 793    Accepted Submission(s): 253


    Problem Description
       Given a map of N * M (2 <= N, M <= 12) , '.' means empty, '*' means walls. You need to build K circuits and no circuits could be nested in another. A circuit is a route connecting adjacent cells in a cell sequence, and also connect the first cell and the last cell. Each cell should be exactly in one circuit. How many ways do we have?

     
    Input
      The first line of input has an integer T, number of cases.
      For each case:
      The first line has three integers N M K, as described above.
      Then the following N lines each has M characters, ‘.’ or ‘*’.
     
    Output
      For each case output one lines.
      Each line is the answer % 1000000007 to the case.
     
    Sample Input
    2
    4 4 1
    **..
    ....
    ....
    ....
     
    4 4 1
    ....
    ....
    ....
    ....
     
    Sample Output
    2
    6
    #include <bits/stdc++.h>
    using namespace std ;
    const int N = 15 ;
    const int M = 30007 ;
    const int MAXN = 1000010;
    const int mod = 1e9+7;
    int n , m , K ;
    int maze[N][N] ;
    int code[N] ;
    int ch[N] , num ;
    int ex , ey ;
    
    struct HASHMAP {
        int head[M] , next[MAXN] , tot  ;
        long long st[MAXN] , f[MAXN] ;
        void init() {
            memset( head , -1 , sizeof head ) ;
            tot = 0 ;
        }
        void push( long long state , long long ans ) {
            int u = state % M ;
            for( int i = head[u] ; ~i ; i = next[i] ) {
                if( st[i] == state ) {
                    f[i] += ans ;
                    f[i] %= mod ;
                    return ;
                }
            }
            st[tot] = state ;
            f[tot] = ans % mod ;
            next[tot] = head[u] ;
            head[u] = tot++ ;
        }
    } mp[2] ;
    
    void decode ( int* code , int m , long long st ) {
        num = st & 63 ;
        st >>= 6 ;
        for( int i = m ; i >= 0 ; --i ) {
            code[i] = st&7 ;
            st >>= 3 ;
        }
    }
    
    long long encode( int *code , int m ) {
        int cnt = 1 ;
        long long st = 0 ;
        memset( ch , -1 , sizeof ch) ;
        ch[0] = 0 ;
        for( int i = 0 ; i <= m ; ++i ) {
            if( ch[code[i]] == -1 ) ch[ code[i] ] = cnt++ ;
            code[i] = ch[ code[i] ] ;
            st <<= 3 ;
            st |= code[i] ;
        }
        st <<= 6 ;
        st |= num ;
        return st ;
    }
    
    void shift( int *code , int m ) {
        for( int i = m ; i > 0 ; --i ) {
            code[i] = code[i-1] ;
        } code[0] = 0 ;
    }
    
    void dpblank( int i , int j , int cur ) {
        int left , up ;
        for( int k = 0 ; k < mp[cur].tot ; ++k ) {
            decode( code , m , mp[cur].st[k] );
            left = code[j-1] ;
            up = code[j] ;
            if( left && up ) {
                if( left == up ) {
                    if( num >= K ) continue ;
                    int c = 0 ;
                    for( int y = 0 ; y < j - 1 ; ++y )
                        if( code[y] ) c++ ;
                    if( c&1 ) continue ;
                    num++ ;
                    code[j-1] = code[j] = 0 ;
                    if( j == m ) shift( code , m ) ;
                    mp[cur^1].push( encode(code,m),mp[cur].f[k] );
                }else {
                    code[j-1] = code[j] = 0 ;
                    for( int t = 0 ; t <= m ; ++t ) {
                        if( code[t] == up )
                            code[t] = left ;
                    }
                    if( j == m ) shift( code,m );
                    mp[cur^1].push(encode(code,m),mp[cur].f[k]) ;
                }
            }
            else if( ( left && ( !up ) ) || ( up && (!left ) ) ) {
                int  t ;
                if( left ) t = left ;
                else t = up ;
                if( maze[i][j+1] ) {
                    code[j-1] = 0 ;
                    code[j] = t ;
                    mp[cur^1].push( encode(code,m) , mp[cur].f[k] ) ;
                }
                if( maze[i+1][j] ) {
                    code[j-1] = t ;
                    code[j] = 0 ;
                    if( j == m ) shift( code , m );
                    mp[cur^1].push(encode(code,m),mp[cur].f[k]);
    
                }
            }
            else {
                if( maze[i][j+1] && maze[i+1][j] ) {
                    code[j-1] = code[j] = 13 ;
                    mp[cur^1].push( encode(code,m),mp[cur].f[k]);
                }
            }
        }
    }
    void dpblock( int i , int j , int cur ) {
        for( int k = 0 ; k < mp[cur].tot ; ++k ) {
            decode( code , m , mp[cur].st[k] );
            code[j-1] = code[j] = 0 ;
            if( j == m ) shift( code , m );
            mp[cur^1].push( encode(code,m) , mp[cur].f[k] );
        }
    }
    
    void Solve() {
        int v = 0 ;
        mp[v].init();
        mp[v].push(0,1);
        for( int i = 1 ; i <= n ; ++i ) {
            for( int j = 1 ; j <= m ; ++j ) {
                mp[v^1].init() ;
                if( maze[i][j] ) dpblank( i , j , v ) ;
                else dpblock( i , j , v );
                v ^= 1 ;
            }
        }
        long long ans = 0 ;
        for( int i = 0 ; i < mp[v].tot ; ++i ) {
           if( mp[v].st[i] == K ) ans = ( ans + mp[v].f[i] ) % mod ;
        }
        cout << ans << endl ;
    }
    string s ;
    
    int main () {
    //    freopen("in.txt","r",stdin);
        ios::sync_with_stdio(0);
        int _ ; cin >> _ ;
        while( _-- ) {
            cin >> n >> m >> K ;
            ex = 0 ;
            memset( maze , 0 , sizeof maze ) ;
            for( int i = 1 ; i <= n ; ++i ) {
                cin >> s ;
                for( int j = 0 ; j < m ; ++j ) {
                    if( s[j] == '.' ) {
                        ex = i , ey = j + 1 ;
                        maze[i][j+1] = 1 ;
                    }
                }
            }
            if( !ex ) { cout << '0' << endl ; continue ; }
            else Solve();
        }
        return 0 ;
    }
    View Code
  • 相关阅读:
    Python模块time、datetime
    Python装饰器、内置函数之金兰契友
    Python生成器、推导式之前襟后裾
    Python函数二(函数名,闭包,迭代器)之杵臼之交
    Python函数(一)之杵臼之交
    katalon系列二:selenium IDE的替代者——Katalon Recorder
    katalon系列一:初识Katalon Studio自动化测试工具
    windows下使用RoboCopy命令进行文件夹增量备份
    Selenium+Python自动化之如何绕过登录验证码
    UI自动化录制工具----UI Recorder
  • 原文地址:https://www.cnblogs.com/hlmark/p/4371353.html
Copyright © 2011-2022 走看看