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HDU 5012 Dice

2014-09-15 19:01 309 查看
有一个六面骰子,有四种翻面方法,求由原状态到目标状态需要翻的最小次数

BFS:

#include <stdio.h>
#include <string.h>
#define MAXN 1024
struct Dice {
int a[6];
int n;
static int dir[4][6];
int value () {
int sum = 0;
for (int i = 0; i < 6; i ++) {
sum *= 10;
sum += a[i];
}
return sum;
}
Dice rotation(int k) {
Dice dice;
for (int i = 0; i < 6; i ++)
dice.a[i] = a[dir[k][i]];
return dice;
}

} queue[MAXN];
int Dice :: dir[4][6] = {{3, 2, 0, 1, 4, 5}, {2, 3, 1, 0, 4, 5}, {5, 4, 2, 3, 0, 1}, {4, 5, 2, 3, 1, 0}};
int values[MAXN], num;
int aim[6];
bool judge (int k) {
for (int i = 0; i < num; i ++)
if (values[i] == k)
return false;
values[num ++] = k;
return true;
}
int BFS () {
int front = 0, rear = 1;
queue[0].n = 0;
while (front < rear) {
Dice& dice = queue[front ++];
if (memcmp(dice.a, aim, sizeof(aim)) == 0)
return dice.n;
for (int i = 0; i < 4; i ++) {
queue[rear] = dice.rotation(i);
if (judge(queue[rear].value())) {
queue[rear ++].n = dice.n + 1;
}
}
}
return -1;
}
int main() {
//freopen ("in.txt", "r", stdin);
while (~scanf ("%d", &queue[0].a[0])) {
num = 0;
for (int i = 1; i < 6; i ++)
scanf ("%d", &queue[0].a[i]);
for (int i = 0; i < 6; i ++) {
scanf ("%d", &aim[i]);
}
printf ("%d\n", BFS());
}
return 0;
}


DFS(由于需要判断骰子可以翻到所有情况(存在大量重复)来确定最小次数,故该方法效率太低,超时):

#include <stdio.h>
#include <string.h>
#define MAXN 1024
typedef int Dice[6];
int dir[4][6] = {{3, 2, 0, 1, 4, 5}, {2, 3, 1, 0, 4, 5}, {5, 4, 2, 3, 0, 1}, {4, 5, 2, 3, 1, 0}};
int values[MAXN], sum;
Dice begin, aim;
bool judge (int k, int num) {
for (int i = 0; i < num; i ++)
if (values[i] == k)
return false;
return true;
}
int value (Dice& a) {
int sum = 0;
for (int i = 0; i < 6; i ++) {
sum *= 10;
sum += a[i];
}
return sum;
}
void DFS (Dice& dice, int cur) {
if (memcmp(dice, aim, sizeof(aim)) == 0 && (cur < sum || sum == -1)) {
sum = cur;
return;
}
Dice ndice;
int k;
for (int i = 0; i < 4; i ++) {
for (int j = 0; j < 6; j ++)
ndice[j] = dice[dir[i][j]];
k = value(ndice);
if (judge(k, cur)) {
values[cur] = k;
DFS(ndice, cur + 1);
}
}
}
int main() {
//freopen ("in.txt", "r", stdin);
while (~scanf ("%d", &begin[0])) {
sum = -1;
for (int i = 1; i < 6; i ++)
scanf ("%d", &begin[i]);
for (int i = 0; i < 6; i ++) {
scanf ("%d", &aim[i]);
}
DFS(begin, 0);
printf ("%d\n", s
4000
um);
}
return 0;
}
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标签:  C++ bfs dfs