2000 Asia shanghai Dance Dance Revolution

思路：dp[i][x][y]表示第i个序列中，右脚在x位置，左脚在y位置时，其最小花费。

那么dp[i][x][y]=min(dp[i-1][a[i]][y]+cost[a[i]][x],dp[i-1][x][a[i]]+cost[a[i]][y]);

题目连接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=32

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<cstdio>
#include<vector>
#include<string>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Maxn 2010
#define Maxm 100010
#define LL __int64
#define Abs(x) ((x)>0?(x):(-x))
#define lson(x) (x<<1)
#define rson(x) (x<<1|1)
#define inf 0x7fffffff
#define Mod 1000000007
using namespace std;
int cost[][]={{,,,,},{,,,,},{,,,,},{,,,,},{,,,,}};
int dp[][][];
int list[];
int main()
{
    int n,i,j,cnt,num,x,y;
    cnt=;
    while(scanf("%d",&list[++cnt]),list[cnt])
    {
        memset(dp,,sizeof(dp));
        while(scanf("%d",&num)){
            if(num!=) {
                list[++cnt]=num;
                continue;
            }
        dp[][list[]][]=cost[][list[]];
        dp[][][list[]]=cost[][list[]];
        for(i=;i<=cnt;i++){
            for(x=;x<=;x++){
                for(y=;y<=;y++){
                    dp[i][list[i]][y]=min(dp[i][list[i]][y],dp[i-][x][y]+cost[x][list[i]]);
                    dp[i][x][list[i]]=min(dp[i][x][list[i]],dp[i-][x][y]+cost[y][list[i]]);
                }
            }
        }
        int ans=inf;
        for(i=;i<=;i++)
            for(j=;j<=;j++)
            ans=min(ans,dp[cnt][i][j]);
            printf("%d\n",ans);
            cnt=;
            break;
        }
    }
    return ;
}