http://codeforces.com/gym/101257/problem/A
把它固定在(0,0, 0)到(2, 2, 2)上,每次都暴力dfs檢查,不會逾時的,因為規定在這個空間上,一不行,就會早早退出。
這樣寫起來比較好寫。
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 100 + 2;
bool vis[4][4][4];
char str[maxn];
int fx[6] = {0, 1, 0, -1, 0, 0};
int fy[6] = {1, 0, -1, 0, 0, 0};
int fz[6] = {0, 0, 0, 0, -1, 1};
bool check(int x, int y, int z) {
if (x < 0 || y < 0 || z < 0) return false;
if (x > 2 || y > 2 || z > 2) return false;
if (vis[x][y][z]) return false;
return true;
}
int op[22];
bool dfs(int cur, int x, int y, int z, int dir) {
int tx = fx[dir] + x, ty = fy[dir] + y, tz = fz[dir] + z;
if (cur == 27) {
if (check(tx, ty, tz)) return true;
else return false;
}
if (!check(tx, ty, tz)) return false;
if (str[cur] == 'I') {
vis[tx][ty][tz] = true;
if (dfs(cur + 1, tx, ty, tz, dir)) {
return true;
} else {
vis[tx][ty][tz] = false;
return false;
}
} else {
vis[tx][ty][tz] = true;
for (int i = 0; i < 6; ++i) {
if (dir == i || op[dir] == i) continue;
if (dfs(cur + 1, tx, ty, tz, i)) {
return true;
}
}
vis[tx][ty][tz] = false;
return false;
}
}
void work() {
op[0] = 2;
op[2] = 0;
op[1] = 3;
op[3] = 1;
op[4] = 5;
op[5] = 4;
scanf("%s", str + 1);
for (int i = 2; i <= 26; ++i) {
if (str[i] == 'E') {
cout << "NO" << endl;
return;
}
}
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 3; ++j) {
for (int k = 0; k < 3; ++k) {
vis[i][j][k] = true;
if (dfs(2, i, j, k, 0)) {
cout << "YES" << endl;
return;
}
vis[i][j][k] = false;
}
}
}
cout << "NO" << endl;
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
work();
return 0;
}