【USACO3.2.5】魔板 康托展开/BFS
2015-03-10 00:24
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我A*然后错了! 其实我不知道我的A*哪里错了…… 但是不管了,以后再练A*好了。
题目思想: 任何一个全排列数字,可以用康拓展开的知识,来映射到一个数字上去。 http://zh.wikipedia.org/wiki/%E5%BA%B7%E6%89%98%E5%B1%95%E5%BC%80
WIKI有详细的讲解。
当然,用散列函数也可以让全排列数字能有映射效果,MAP什么都可以解决这个问题……
其实就是BFS啦! 核心问题就是,判断这个状态是否进过队列。 把这个8位全排列,用康拓展开映射成数字,或者直接散列函数,或者扔进平衡树,都可以解决问题……
我的程序因为原来想写A*,然后写挂了……所以很丑很长……而且也不快
/*
TASK:msquare
LANG:C++
*/
#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
int factorial[10]={1,1,2,6,24,120,720,5040,40320,362880};
bool vis[10];
int pnext[10], ppre[10];
inline void clean()
{
for (int i = 1; i <= 8; ++ i)
{
ppre[i] = i - 1;
pnext[i] = i + 1;
}
pnext[0] = 1;
ppre[9] = 9;
}
inline void del(int k)
{
pnext[ppre[k]] = pnext[k];
ppre[pnext[k]] = ppre[k];
}
inline int cantor_expand(int *r, int n = 8)//r数组(含0下标) 有n位,转化为一个数字
{
int ans = 0;
clean();
for (int i = 0; i != n; ++ i)
{
int tmp = -1;
for (int j = 0; j < r[i]; j = pnext[j], ++ tmp);
del(r[i]);
//cout<<tmp<<" " << n - i - 1 << endl;
ans += tmp * factorial[n - i - 1];
}
return ans;
}
inline void cantor_expand(int k, int *r, int n = 8)//k这个数字,有n位,转化进r数组
{
clean();
for (int i = 0; i != n - 1; ++ i)
{
int tmp = k / factorial[n - i - 1], count = 0, j;
for (j = pnext[0]; ; j = pnext[j])
{
++ count;
if (count == tmp + 1) break;
}
r[i] = j;
del(j);
k %= factorial[n - i - 1];
}
r[n - 1] = pnext[0];
}
//init 函数:处理最短路径问题
int g[9][9];//保存每个数字到每个地方的距离
void init()
{
int num[2][4],t=0;
for (int i = 0; i <= 1; ++ i)
for (int j = 0; j != 4; ++ j) num[i][j] = t ++ ;
memset(g, 40, sizeof(g));
int inf = g[0][0];
for (int i = 0; i != 9; ++ i) g[i][i] = 0;
g[1][8] = g[2][7] = g[3][6] = g[4][5] = 1;
g[8][1] = g[7][2] = g[6][3] = g[5][4] = 1;
g[1][2] = g[2][3] = g[3][4] = g[4][1] = 1;
g[8][7] = g[7][6] = g[6][5] = g[5][8] = 1;
g[6][7] = 1;
for (int k = 0; k != 9; ++ k)
for (int i = 0; i != 9; ++ i)
for (int j = 0; j != 9; ++ j)
if (g[i][k] + g[k][j] < g[i][j]) g[i][j] = g[i][k] + g[k][j];
}
int cantor[10];
struct node
{
int way;
int did, willdid;
int num;
node(int W, int D, int NUM):way(W),did(D),num(NUM)
{
cantor_expand(num, cantor);
//for (int i = 0; i != 8; ++ i) cout<<cantor[i]<<" ";cout<<endl;
willdid = 0;
for (int i = 0; i != 8; ++ i)
{
int tmp = g[cantor[i]][i + 1];
willdid = max(willdid, tmp);
}
willdid = 0;
}
};
inline bool operator < (node A, node B)
{
if (A.did + A.willdid == B.did + B.willdid) return A.way > B.way; //估价步数相同,使用转移
return A.did + A.willdid > B.did + B.willdid;
}
bool v[41000] = {0};
int pre[41000]={0}, plan[41000];
int cantor_tmp[10];
//priority_queue<node>q;
queue<node>q;
inline int one(int now, int did)
{
for (int i = 0; i != 8; ++ i) cantor[i] = cantor_tmp[i];
swap(cantor[0], cantor[7]);
swap(cantor[1], cantor[6]);
swap(cantor[2], cantor[5]);
swap(cantor[3], cantor[4]);
int tmp = cantor_expand(cantor);
if (!v[tmp])
{
v[tmp] = 1;
pre[tmp] = now;
plan[tmp] = 1;
q.push(node(1, did, tmp));
}
return tmp;
}
inline int two(int now, int did)
{
for (int i = 0; i != 8; ++ i) cantor[i] = cantor_tmp[i];
int tmp = cantor[3];
for (int i = 3; i >= 1; -- i) cantor[i] = cantor[i - 1];
cantor[0] = tmp;
tmp = cantor[4];
for (int i = 4; i <= 7; ++ i) cantor[i] = cantor[i + 1];
cantor[7] = tmp;
tmp = cantor_expand(cantor);
if (!v[tmp])
{
v[tmp] = 1;
pre[tmp] = now;
plan[tmp] = 2;
q.push(node(2, did, tmp));
}
return tmp;
}
inline int three(int now, int did)
{
for (int i = 0; i != 8; ++ i) cantor[i] = cantor_tmp[i];
int tmp = cantor[6];
cantor[6] = cantor[5];
cantor[5] = cantor[2];
cantor[2] = cantor[1];
cantor[1] = tmp;
tmp = cantor_expand(cantor);
if (!v[tmp])
{
v[tmp] = 1;
pre[tmp] = now;
plan[tmp] = 3;
q.push(node(3, did, tmp));
}
return tmp;
}
int output[41000],ot=0;
void doit()
{
int aim, tmp, begin(0), a[10];
for (int i = 0; i != 8; ++ i) cin >> a[i];
aim = cantor_expand(a);
q.push(node(0, 0, begin));
v[begin] = 1;
while (1)
{
int now = q.front().num;
int did = q.front().did;
q.pop();
cantor_expand(now, cantor_tmp);
if (one(now, did + 1) == aim) break;
if (two(now, did + 1) == aim) break;
if (three(now, did + 1) == aim) break;
}
int now = aim;
while (now != begin)
{
output[ot++] = plan[now];
now = pre[now];
}
cout<<ot<<endl;
for (int i = ot - 1; i >= 0; -- i) cout<<(char)(output[i]+'A' - 1);
cout<<endl;
}
int main()
{
freopen("msquare.in","r",stdin);
freopen("msquare.out","w",stdout);
init();
doit();
return 0;
}
题目思想: 任何一个全排列数字,可以用康拓展开的知识,来映射到一个数字上去。 http://zh.wikipedia.org/wiki/%E5%BA%B7%E6%89%98%E5%B1%95%E5%BC%80
WIKI有详细的讲解。
当然,用散列函数也可以让全排列数字能有映射效果,MAP什么都可以解决这个问题……
其实就是BFS啦! 核心问题就是,判断这个状态是否进过队列。 把这个8位全排列,用康拓展开映射成数字,或者直接散列函数,或者扔进平衡树,都可以解决问题……
我的程序因为原来想写A*,然后写挂了……所以很丑很长……而且也不快
Compiling... Compile: OK Executing... Test 1: TEST OK [0.005 secs, 4024 KB] Test 2: TEST OK [0.003 secs, 4024 KB] Test 3: TEST OK [0.003 secs, 4024 KB] Test 4: TEST OK [0.003 secs, 4024 KB] Test 5: TEST OK [0.022 secs, 4024 KB] Test 6: TEST OK [0.035 secs, 4024 KB] Test 7: TEST OK [0.051 secs, 4024 KB] Test 8: TEST OK [0.073 secs, 4024 KB] All tests OK.
/*
TASK:msquare
LANG:C++
*/
#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
int factorial[10]={1,1,2,6,24,120,720,5040,40320,362880};
bool vis[10];
int pnext[10], ppre[10];
inline void clean()
{
for (int i = 1; i <= 8; ++ i)
{
ppre[i] = i - 1;
pnext[i] = i + 1;
}
pnext[0] = 1;
ppre[9] = 9;
}
inline void del(int k)
{
pnext[ppre[k]] = pnext[k];
ppre[pnext[k]] = ppre[k];
}
inline int cantor_expand(int *r, int n = 8)//r数组(含0下标) 有n位,转化为一个数字
{
int ans = 0;
clean();
for (int i = 0; i != n; ++ i)
{
int tmp = -1;
for (int j = 0; j < r[i]; j = pnext[j], ++ tmp);
del(r[i]);
//cout<<tmp<<" " << n - i - 1 << endl;
ans += tmp * factorial[n - i - 1];
}
return ans;
}
inline void cantor_expand(int k, int *r, int n = 8)//k这个数字,有n位,转化进r数组
{
clean();
for (int i = 0; i != n - 1; ++ i)
{
int tmp = k / factorial[n - i - 1], count = 0, j;
for (j = pnext[0]; ; j = pnext[j])
{
++ count;
if (count == tmp + 1) break;
}
r[i] = j;
del(j);
k %= factorial[n - i - 1];
}
r[n - 1] = pnext[0];
}
//init 函数:处理最短路径问题
int g[9][9];//保存每个数字到每个地方的距离
void init()
{
int num[2][4],t=0;
for (int i = 0; i <= 1; ++ i)
for (int j = 0; j != 4; ++ j) num[i][j] = t ++ ;
memset(g, 40, sizeof(g));
int inf = g[0][0];
for (int i = 0; i != 9; ++ i) g[i][i] = 0;
g[1][8] = g[2][7] = g[3][6] = g[4][5] = 1;
g[8][1] = g[7][2] = g[6][3] = g[5][4] = 1;
g[1][2] = g[2][3] = g[3][4] = g[4][1] = 1;
g[8][7] = g[7][6] = g[6][5] = g[5][8] = 1;
g[6][7] = 1;
for (int k = 0; k != 9; ++ k)
for (int i = 0; i != 9; ++ i)
for (int j = 0; j != 9; ++ j)
if (g[i][k] + g[k][j] < g[i][j]) g[i][j] = g[i][k] + g[k][j];
}
int cantor[10];
struct node
{
int way;
int did, willdid;
int num;
node(int W, int D, int NUM):way(W),did(D),num(NUM)
{
cantor_expand(num, cantor);
//for (int i = 0; i != 8; ++ i) cout<<cantor[i]<<" ";cout<<endl;
willdid = 0;
for (int i = 0; i != 8; ++ i)
{
int tmp = g[cantor[i]][i + 1];
willdid = max(willdid, tmp);
}
willdid = 0;
}
};
inline bool operator < (node A, node B)
{
if (A.did + A.willdid == B.did + B.willdid) return A.way > B.way; //估价步数相同,使用转移
return A.did + A.willdid > B.did + B.willdid;
}
bool v[41000] = {0};
int pre[41000]={0}, plan[41000];
int cantor_tmp[10];
//priority_queue<node>q;
queue<node>q;
inline int one(int now, int did)
{
for (int i = 0; i != 8; ++ i) cantor[i] = cantor_tmp[i];
swap(cantor[0], cantor[7]);
swap(cantor[1], cantor[6]);
swap(cantor[2], cantor[5]);
swap(cantor[3], cantor[4]);
int tmp = cantor_expand(cantor);
if (!v[tmp])
{
v[tmp] = 1;
pre[tmp] = now;
plan[tmp] = 1;
q.push(node(1, did, tmp));
}
return tmp;
}
inline int two(int now, int did)
{
for (int i = 0; i != 8; ++ i) cantor[i] = cantor_tmp[i];
int tmp = cantor[3];
for (int i = 3; i >= 1; -- i) cantor[i] = cantor[i - 1];
cantor[0] = tmp;
tmp = cantor[4];
for (int i = 4; i <= 7; ++ i) cantor[i] = cantor[i + 1];
cantor[7] = tmp;
tmp = cantor_expand(cantor);
if (!v[tmp])
{
v[tmp] = 1;
pre[tmp] = now;
plan[tmp] = 2;
q.push(node(2, did, tmp));
}
return tmp;
}
inline int three(int now, int did)
{
for (int i = 0; i != 8; ++ i) cantor[i] = cantor_tmp[i];
int tmp = cantor[6];
cantor[6] = cantor[5];
cantor[5] = cantor[2];
cantor[2] = cantor[1];
cantor[1] = tmp;
tmp = cantor_expand(cantor);
if (!v[tmp])
{
v[tmp] = 1;
pre[tmp] = now;
plan[tmp] = 3;
q.push(node(3, did, tmp));
}
return tmp;
}
int output[41000],ot=0;
void doit()
{
int aim, tmp, begin(0), a[10];
for (int i = 0; i != 8; ++ i) cin >> a[i];
aim = cantor_expand(a);
q.push(node(0, 0, begin));
v[begin] = 1;
while (1)
{
int now = q.front().num;
int did = q.front().did;
q.pop();
cantor_expand(now, cantor_tmp);
if (one(now, did + 1) == aim) break;
if (two(now, did + 1) == aim) break;
if (three(now, did + 1) == aim) break;
}
int now = aim;
while (now != begin)
{
output[ot++] = plan[now];
now = pre[now];
}
cout<<ot<<endl;
for (int i = ot - 1; i >= 0; -- i) cout<<(char)(output[i]+'A' - 1);
cout<<endl;
}
int main()
{
freopen("msquare.in","r",stdin);
freopen("msquare.out","w",stdout);
init();
doit();
return 0;
}
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