UVALive

题目大意：有一个胖子在玩跳舞机。刚開始的位置在(0,0)。跳舞机有四个方向键，上左下右分别相应1,2,3,4.如今有下面规则
1.假设从0位置移动到随意四个位置，消耗能量2
2.假设从非0位置跳到相邻的位置，如1跳到2或4，消耗能量3
3.假设从非0位置跳到对面的位置。如2跳到4。消耗能量4
4.假设跳同一个位置，消耗能量1
5.两仅仅脚不能在同一个位置

解题思路：这题事实上非常水。直接暴力就能够攻克了，讨论全部情况，用dp[i][j][k]表示跳第k个数字。左脚在i这个位置。右脚在j这个位置时所消耗的能量，接着分类讨论

1.假设当中一仅仅脚在0上的情况
2.当中一仅仅脚踩的数字和当前要跳的数字一样
3.两仅仅脚踩的数字和当前的数字不一样

三种情况，分别在细分就可以，详细看代码

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define maxn 50010
#define INF 0x3f3f3f3f
int dp[5][5][maxn];
int seq[maxn];
int strength[2] = {4,3};
int n;

int solve() {
    memset(dp, 0x3f, sizeof(dp));
    dp[0][seq[0]][0] = dp[seq[0]][0][0] = 2;

    for(int i = 1; i < n; i++) {
        for(int j = 0; j < 5; j++) {
            if(dp[j][seq[i-1]][i-1] != INF) {

                if(j == 0) {
                    if(seq[i] != seq[i-1])
                        dp[seq[i]][seq[i-1]][i] = dp[j][seq[i-1]][i-1] + 2;

                    if(seq[i] == seq[i-1])
                        dp[j][seq[i-1]][i] = dp[j][seq[i-1]][i-1] + 1;
                    else
                        dp[j][seq[i]][i] = dp[j][seq[i-1]][i-1] + strength[(seq[i-1] + seq[i]) % 2];    
                }
                else if(j == seq[i] || seq[i-1] == seq[i]) 
                    dp[j][seq[i-1]][i] = min(dp[j][seq[i-1]][i],dp[j][seq[i-1]][i-1] + 1);
                else {
                    dp[seq[i]][seq[i-1]][i] = min(dp[j][seq[i-1]][i-1] + strength[(j + seq[i]) % 2], dp[seq[i]][seq[i-1]][i]);
                    dp[j][seq[i]][i] = min(dp[j][seq[i-1]][i-1] + strength[(seq[i-1] + seq[i] ) % 2], dp[j][seq[i]][i]);
                }
            }

            if(dp[seq[i-1]][j][i-1] != INF) {
                if(j == 0) {
                    if(seq[i] != seq[i-1])
                        dp[seq[i]][seq[i-1]][i] = dp[seq[i-1]][j][i-1] + 2;

                    if(seq[i] == seq[i-1])
                        dp[seq[i-1]][j][i] = dp[seq[i-1]][j][i-1] + 1;
                    else
                        dp[seq[i]][j][i] = dp[seq[i-1]][j][i-1] + strength[(seq[i-1] + seq[i]) % 2];    
                }

                if(j == seq[i] || seq[i-1] == seq[i]) 
                    dp[seq[i-1]][j][i] = min(dp[seq[i-1]][j][i],dp[seq[i-1]][j][i-1] + 1);
                else {
                    dp[seq[i]][seq[i-1]][i] = min(dp[seq[i-1]][j][i-1] + strength[(j + seq[i]) % 2], dp[seq[i]][seq[i-1]][i]);
                    dp[seq[i]][j][i] = min(dp[seq[i-1]][j][i-1] + strength[(seq[i-1] + seq[i] ) % 2], dp[seq[i]][j][i]);
                }
            }
        }
    }
    int ans = INF;
    for(int i = 0; i < 5; i++)
        ans = min(min(ans, dp[seq[n-1]][i][n-1]), dp[i][seq[n-1]][n-1]);

    return ans;
}

int main() {
    n = 0;
    while(scanf("%d", &seq[n]) != EOF && seq[n++]) {
        while(scanf("%d", &seq[n]) && seq[n])
            n++;    
        printf("%d
", solve());
        n = 0;
    }
    return 0;
}