uva 10051 Tower of Cubes(DAG最长路）
题目连接：10051 - Tower of Cubes
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\n", Max); for (int i = 0; i < Max; i++) {printf("%d %s\n", idx, sign[idy]);a = idx;idx = x[idx][idy];idy = y[a][idy]; } /* printf("%d\n", dp[1][5]); idx = 1; idy = 5; for (int i = 0; idx; i++) {printf("%d %s\n", 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("\n");printf("Case #%d\n", ++cas);solve(); } return 0;}