题目大意:有n个正方体,从序号1~n, 对应的每个立方体的6个面分别有它的颜色(用数字给出),现在想要将立方体堆成塔,并且上面的立方体的序号要小于下面立方体的序号,相邻的面颜色必须相同。输出最高值和路径。
解题思路:因为立方体可以旋转,所以一个序号的立方体对应这6种不同的摆放方式,可以将问题理解成DAG最长路问题, 只是搜索范围是从i + 1开始到n。然后记录路径要开两个2维数组。
路径不唯一,随便输出一条。
#include <stdio.h> #include <string.h> const int N = 1005; const int M = 10; const char sign[M][10]= {"front", "back", "left", "right", "top", "bottom"}; struct state { int in; int out; }tmp[N][M]; int n, x[N][M], y[N][M], dp[N][M]; void Init() { memset(tmp, 0, sizeof(tmp)); memset(dp, 0, sizeof(dp)); memset(x, 0, sizeof(x)); memset(y, 0, sizeof(y)); } void write(int k, int a, int b, int d) { tmp[d][k].in = a; tmp[d][k].out = b; } void read() { int a, b; for (int i = 1; i <= n; i++) { for (int j = 0; j < 3; j++) { scanf("%d%d", &a, &b); write(j * 2, a, b, i); write(j * 2 + 1, b, a, i); } } } int search(int d, int k) { if (dp[d][k]) return dp[d][k]; for (int i = d + 1; i <= n; i++) { for (int j = 0; j < 6; j++) { if (tmp[i][j].in == tmp[d][k].out) { int a = search(i, j); if (a > dp[d][k]) { dp[d][k] = a; x[d][k] = i, y[d][k] = j; } } } } return ++dp[d][k]; } void solve() { int Max = 0, idx, idy, a; for (int i = 1; i <= n; i++) { if (Max + i >= n) break; for (int j = 0; j < 6; j++) { a = search(i, j); if (a > Max) { Max = a; idx = i, idy = j; } } } printf("%d ", Max); for (int i = 0; i < Max; i++) { printf("%d %s ", idx, sign[idy]); a = idx; idx = x[idx][idy]; idy = y[a][idy]; } /* printf("%d ", dp[1][5]); idx = 1; idy = 5; for (int i = 0; idx; i++) { printf("%d %s ", idx, sign[idy]); a = idx; idx = x[idx][idy]; idy = y[a][idy]; } */ } int main() { int cas = 0; while (scanf("%d", &n), n) { Init(); read(); if (cas) printf(" "); printf("Case #%d ", ++cas); solve(); } return 0; }