USACO 3.2.5

纠结了几天，在昨晚AC了一道bfs后，又鏖战了几个小时，学了很多新东西，写了140+行代码，终于AC.

题目的本质是一道bfs，状态8!=40320(不是7!,顺便就在NOCOW上纠正了)，不需要剪枝（一个简单的剪枝是到达目标状态就停止）.

个人认为难点有两个:

1.节点编解码要用到康托展开;

2.答案要求输出操作序列，所以要记录方案.

另外要注意操作数为0的情况，以及USACO的linux系统与Windows的区别.

(康托展开和输出操作序列的方法将另行记录)

 

另:关于状态数

7位hash没问题，但是空间仍是8!=40320，而不是7!=5040.因为如果空间是7!的话，那么排列1234567和1234568将算作一个排列，重复了，所以一定会错. 另外，如果使用Contor展开的话，必须是8位，因为Contor展开是基于全排列的.

--by Climber.pI

Executing...
   Test 1: TEST OK [0.011 secs, 3688 KB]
   Test 2: TEST OK [0.011 secs, 3688 KB]
   Test 3: TEST OK [0.000 secs, 3688 KB]
   Test 4: TEST OK [0.000 secs, 3688 KB]
   Test 5: TEST OK [0.011 secs, 3688 KB]
   Test 6: TEST OK [0.043 secs, 3688 KB]
   Test 7: TEST OK [0.054 secs, 3688 KB]
   Test 8: TEST OK [0.086 secs, 3688 KB]

All tests OK.

Your program ('msquare') produced all correct answers!  This is your
submission #4 for this problem.  Congratulations!

 

[Code](由于Sina的限制就没代码高亮了)优快云自带代码高亮.
#include<stdio.h> #include<string.h> #include<iostream> using namespace std; struct square{int b[8];}; const int MAXN = 40320+10; bool vis[MAXN] = {0}; int dist[MAXN] = {0}, fa[MAXN] = {0}, op[MAXN] = {0}; square aim, start; FILE *fin, *fout; struct queue { int f, r, k, n, q[MAXN]; queue() {memset(q, 0, sizeof(q)); n = k = f= r = 0;} void push(int x){q[k++] = x; r++; n++;} int pop(){n--; return q[f++];} bool empty(){return (f<r) ? 0 : 1;} }; queue q; int contor(square& t)//康托展开 { int ans = 0, frac[8]= {1, 1, 2, 6, 24, 120, 720, 5040}; for (int i = 0; i < 8; i++) { int tmp = 0; for (int j = i+1; j < 8; j++) if (t.b[i] > t.b[j]) ++tmp; ans += tmp * frac[7-i]; } return ans+1; } void uncontor(int p, square& ans)//逆康托展开 { if (p < 0) return; p--; int frac[]= {1, 1, 2, 6, 24, 120, 720, 5040, 40320}; bool hash[8] = {0}; for (int i = 0; i < 8 ; i++) { int tmp = p / frac[7-i]; for (int j = 0; j <= tmp; j++) if(hash[j]) tmp++; ans.b[i] = tmp+1; hash[tmp] = 1; p %= frac[7-i]; } } void change(int p, square& a) { int tmp; switch (p) { case 1: for (int i = 0; i < 4; i++) { tmp = a.b[i]; a.b[i] = a.b[7-i]; a.b[7-i] = tmp; } break; case 2: tmp = a.b[3]; for (int i = 3; i > 0; i--) a.b[i] = a.b[i-1]; a.b[0] = tmp; tmp = a.b[4]; for (int i = 4; i < 7; i++) a.b[i] = a.b[i+1]; a.b[7] = tmp; break; case 3: tmp = a.b[1]; a.b[1] = a.b[6]; a.b[6] = a.b[5]; a.b[5] = a.b[2]; a.b[2] = tmp; break; } } bool check(square& a) { for (int i = 0; i < 8; i++) if (a.b[i] != aim.b[i]) return false; return true; } void bfs() { int node = contor(start); q.push(node); fa[node] = node; vis[node] = 1; dist[node] = 0; while(!q.empty()) { node = q.pop(); square unnode; uncontor(node, unnode); for (int i = 1; i < 4; i++) { square tmp; memcpy(tmp.b, unnode.b, sizeof(unnode.b)); change(i, tmp); int _node = contor(tmp); if (!vis[_node]) { q.push(_node); vis[_node] = 1; fa[_node] = node; op[_node] = i; dist[_node] = dist[node]+1; if (check(tmp)) return; } } } } void print(int node)//输出操作 { int c = 0; square tmp; int dir[30]; for(;;) { uncontor(node, tmp); if(node == 1) break; dir[c++] = op[node]; node = fa[node]; } while(c--) fputc(dir[c]+'A'-1, fout); } int main() { fin = fopen("msquare.in", "r"); fout = fopen("msquare.out", "w"); for (int i = 0; i < 8; i++) { fscanf(fin, "%d", &aim.b[i]); start.b[i] = i+1; } if (check(start)) fprintf(fout, "0/n/n"); else { bfs(); fprintf(fout, "%d/n", dist[contor(aim)]); print(contor(aim)); fprintf(fout, "/n"); } }