USACO6.1.1 Postal Vans (vans) 插头dp

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题意：一个4*n的矩阵，从左上角出发，经过每个顶点恰好1次回到左上角，问有多少种走法。

第一道插头dp……磕磕绊绊总算过了……

什么是插头dp请参考cdq的论文《基于连通性状态压缩的动态规划问题》。

注意同样的路径，正着走和反着走算两条不同的，所以最后答案要乘2。

然后加上高精就行了。

代码（括号表示法）：

/*ID:shijiey1LANG:C++PROG:vans*/#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int n = 4;struct Num{short arr[500];int len;void init(int i) {memset(arr, 0, sizeof(arr));arr[0] = i;len = 1;}void print() {for (int i = len - 1; i >= 0; i--) {cout << arr[i];}cout << endl;}};void operator += (Num &a, Num b) {a.len = max(a.len, b.len);for (int i = 0; i < a.len; i++) {a.arr[i] += b.arr[i];a.arr[i + 1] += a.arr[i] / 10;a.arr[i] %= 10;}if (a.arr[a.len]) a.len++;}int N;int stat[1110];int brk[1110][8], stack[8], top = 0, tot = 0;Num dp[8][1110];int main() {freopen("vans.in", "r", stdin);freopen("vans.out", "w", stdout);cin >> N;int m = 1 << ((n + 1) << 1);for (int i = 0; i < m; i++) {top = 0;for (int j = 0; j <= n; j++) {int x = i >> (j << 1);if ((x & 3) == 1) stack[top++] = j;if ((x & 3) == 2)if (top--) {brk[tot][j] = stack[top];brk[tot][stack[top]] = j;} else break;if ((x & 3) == 3) {top = -1;break;}}if (!top) stat[tot++] = i;}Num ans;ans.init(0);memset(dp, 0, sizeof(dp));dp[n][0].init(1);for (int k = 1; k <= N; k++) {for (int i = 0; i < tot; i++) {if (stat[i] & 3) dp[0][stat[i]].init(0);else dp[0][stat[i]] = dp[n][stat[i] >> 2];}for (int i = 1; i <= n; i++) {int x = (i - 1) << 1;memset(dp[i], 0, sizeof(dp[i]));for (int j = 0; j < tot; j++) {int p = (stat[j] >> x) & 3;int q = (stat[j] >> (x + 2)) & 3;// ## -> ()// 9 = (21)4if (!p && !q) dp[i][stat[j] | (9 << x)] += dp[i - 1][stat[j]];else if (p && q) {if (p == q) {// ((...)) -> // ##...()// 5 = (11)4 : ## = (( ^ 5// () = )) ^ 3if (p == 1) dp[i][stat[j] ^ (5 << x) ^ (3 << (brk[j][i] << 1))] += dp[i - 1][stat[j]];// ((...)) ->// ()...##// 10 = (22)4else dp[i][stat[j] ^ (10 << x) ^ (3 << (brk[j][i - 1] << 1))] += dp[i - 1][stat[j]];} else {if (p == 1) {if (k == N && i == n && stat[j] == (9 << x)) ans += dp[i - 1][stat[j]];} else dp[i][stat[j] ^ (6 << x)] += dp[i - 1][stat[j]]; // )( -> ##, 6 = (12)4}} else {dp[i][stat[j]] += dp[i - 1][stat[j]];// #( -> (#dp[i][stat[j] ^ (p << x) ^ (q << x + 2) | (p << x + 2) | (q << x)] += dp[i - 1][stat[j]];}}}}ans += ans;ans.print();return 0;}

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