UVALive 2031 Dance Dance Revolution

题解:

简单DP

dp[i][j][k] 表示第i步双脚在位置j和位置k的位置

然后根据题意推一下转移方程就行了

代码:

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define pii pair<int,int>
#define ll long long

const int INF = 1e9;
const int maxn = 50050;

vector< int > v;
int x;
int dp[ maxn ][ 5 ][ 5 ];

int step_num( int s, int e )
{
    if( s == 0 ) return 2;
    if( s == 1 )
    {
        if( e == 1 ) return 1;
        if( e == 2 || e == 4 ) return 3;
        if( e == 3 ) return 4;
    }
    if( s == 2 )
    {
        if( e == 2 ) return 1;
        if( e == 1 || e == 3 ) return 3;
        if( e == 4 ) return 4;
    }
    if( s == 3 )
    {
        if( e == 3 ) return 1;
        if( e == 2 || e == 4 ) return 3;
        if( e == 1 ) return 4;
    }
    if( s == 4 )
    {
        if( e == 4 ) return 1;
        if( e == 1 || e == 3 ) return 3;
        if( e == 2 ) return 4;
    }
}

int main()
{
    while( ~scanf( "%d", &x ) && x )
    {
        v.clear();
        v.push_back( x );
        while( scanf( "%d", &x ) && x ) v.push_back( x );
        for( int i = 0; i < 5; i ++ )
        for( int j = 0; j < 5; j ++ )
        for( int k = 0; k < 50000; k ++ ) dp[ k ][ i ][ j ] = INF;
        int n = v.size(), goal = INF;
        dp[ 1 ][ 0 ][ v[ 0 ] ] = dp[ 1 ][ v[ 0 ] ][ 0 ] = 2;
        for( int i = 2; i <= n; i ++ )
        {
            int now = v[ i - 1 ];
            for( int j = 0; j < 5; j ++ )
            for( int k = 0; k < 5; k ++ )
            {
                if( j == k ) continue;
                if( j != now ) dp[ i ][ j ][ now ] = min( dp[ i ][ j ][ now ], dp[ i - 1 ][ j ][ k ] + step_num( k, now ) );
                if( k != now ) dp[ i ][ now ][ k ] = min( dp[ i ][ now ][ k ], dp[ i - 1 ][ j ][ k ] + step_num( j, now ) );
                if( i == n )
                {
                    goal = min( goal, dp[ i ][ now ][ k ] );
                    goal = min( goal, dp[ i ][ j ][ now ] );
                }
            }
        }
        printf( "%d\n", goal );
    }
    return 0;
}