uva 10051

将每一个分解为六个两面的 简单地dp 回溯输出路径.....

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
struct cc
{
    int top,bom,fa,w;
    void f(int a, int b, int c, int d)
    {
        top = a;
        bom = b;
        fa = c;
        w = d;
    }
};
cc cube[3100];
char face[7][10] = { "front","back", "left", "right", "top", "bottom"};
int dp[3100],cu[7],p[3100];
void printpath(int k)
{
    if(k == -1)
        return ;
    printpath(p[k]);
    printf("%d %s\n",cube[k].w, face[cube[k].fa]);
}
int main()
{
    int n, ca = 1;
    while(scanf("%d", &n) == 1 && n)
    {
        int m = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < 6; j++)
                scanf("%d",&cu[j]);
            for(int j = 0; j < 6; j++)
            {
                cube[m++].f(cu[j], j%2 == 1?cu[j-1]:cu[j+1], j, i+1);
            }
        }
        memset(dp, 0, sizeof( dp));
        memset(p, -1, sizeof( p));
        int _max = 0, k;
        for(int i = 0; i < m; i++)
            for(int j = i+1; j < m; j++)
                if(cube[j].w > cube[i].w && cube[i].bom == cube[j].top && dp[j] < dp[i]+1)
                {
                    dp[j] = dp[i]+1;
                    p[j] = i;
                    if(dp[j] > _max)
                    {
                        _max = dp[j];
                        k = j;
                    }
                }
        printf("Case #%d\n%d\n",ca++,_max+1);
        printpath(k);
        puts("");
    }
    return 0;
}


posted @ 2013-08-03 14:39  xlc2845  阅读(137)  评论(0编辑  收藏  举报