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;}