UVa 1291 Dance Dance Revolution

dp[k][i][j]表示第k步左脚在位置 i，右脚在位置 j 状态时的最小能量消耗.

dp[k][i][j] = min( dp[ k - 1 ][x][j] + cost[x][ step[k] ], dp[ k - 1 ][i][x] + cost[x][ step[k] ] );

这题很坑爹的没有数据范围，数组我随便开的，居然1Y了……

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>

using namespace std;

const int MAXN = 10010;
const int INF = 1 << 30;

int step[MAXN];
int dp[MAXN][6][6];
int cost[6][6];
int N;

void init()
{
    for ( int i = 0; i < 5; ++i ) cost[i][i] = 1;
    for ( int i = 1; i < 5; ++i ) cost[0][i] = 2;
    for ( int i = 1; i < 5; ++i )
    {
        cost[i][i + 1] = 3;
        cost[i + 1][i] = 3;
    }
    cost[1][4] = 3;
    cost[4][1] = 3;
    cost[1][3] = 4;
    cost[3][1] = 4;
    cost[2][4] = 4;
    cost[4][2] = 4;

    return;
}

void DP()
{
    dp[0][0][0] = 0;
    for ( int k = 1; k < N; ++k )
    {
        for ( int i = 0; i <= 4; ++i )
        {
            int j = step[k];
            for ( int x = 0; x <= 4; ++x )
            {
                dp[k][j][i] = min( dp[k][j][i], dp[k - 1][x][i] + cost[x][j] );
                dp[k][i][j] = min( dp[k][i][j], dp[k - 1][i][x] + cost[x][j] );
            }
        }
    }

    int ans = INF;
    for ( int i = 0; i <= 4; ++i )
    for ( int j = 0; j <= 4; ++j )
        ans = min( ans, dp[N - 1][i][j] );

    printf( "%d\n", ans );

    return;
}

int main()
{
    init();
    int a;
    while( scanf( "%d", &a ), a )
    {
        step[1] = a;
        N = 1;
        while ( scanf( "%d", &step[ ++N ] ), step[N] );

        for ( int i = 0; i <= N; ++i )
            for ( int j = 0; j <= 4; ++j )
                for ( int k = 0; k <= 4; ++k )
                    dp[i][j][k] = INF;

        DP();
    }
    return 0;
}