POJ_1739
即便看上去这个题目不像一个回路问题，但由于规定了起点和中间，实际上我们只要把起点和终点连一条线，这个题目就算变成一个求回路的问题了，只不过这个回路有些特殊。
具体处理的时候可以从下向上dp，然后把初始状态设置成在起点和终点的位置有两个相同的上插头即可。
#include#include #define HASH 30007#define MAXD 15#define SIZE 1000010int N, M, maze[MAXD][MAXD], code[MAXD], ch[MAXD], ex, ey;char b[MAXD];struct Hashmap{ int head[HASH], next[SIZE], state[SIZE], size; long long f[SIZE]; void init() { memset(head, -1, sizeof(head)); size = 0; } void push(int st, long long ans) { int i, h = st % HASH; for(i = head[h]; i != -1; i = next[i]) if(st == state[i]) { f[i] += ans; return ; } state[size] = st, f[size] = ans; next[size] = head[h]; head[h] = size ++; }}hm[2];void decode(int *code, int m, int st){ int i; for(i = m; i >= 0; i --) { code[i] = st & 7; st >>= 3; }}int encode(int *code, int m){ int i, cnt = 1, st = 0; memset(ch, -1, sizeof(ch)); ch[0] = 0; for(i = 0; i <= m; i ++) { if(ch[code[i]] == -1) ch[code[i]] = cnt ++; code[i] = ch[code[i]]; st <<= 3; st |= code[i]; } return st;}void init(){ int i, j, k; ex = 0; memset(maze, 0, sizeof(maze)); for(i = 1; i <= N; i ++) { scanf("%s", b + 1); for(j = M; j >= 1; j --) if(b[j] == '.') { maze[i][j] = 1; if(ex == 0) ex = i, ey = j; } }}void shift(int *code, int m){ int i; for(i = m; i > 0; i --) code[i] = code[i - 1]; code[0] = 0;}void dpblank(int i, int j, int cur){ int k, t, left, down; for(k = 0; k < hm[cur].size; k ++) { decode(code, M, hm[cur].state[k]); left = code[j - 1], down = code[j]; if(left && down) { if(left == down) { if(i == ex && j == ey) { code[j - 1] = code[j] = 0; if(j == M) shift(code, M); hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]); } } else { code[j - 1] = code[j] = 0; for(t = 0; t <= M; t ++) if(code[t] == down) code[t] = left; if(j == M) shift(code, M); hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]); } } else if(left || down) { if(maze[i][j + 1]) { code[j - 1] = 0, code[j] = left + down; hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]); } if(maze[i - 1][j]) { code[j - 1] = left + down, code[j] = 0; if(j == M) shift(code, M); hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]); } } else { if(maze[i - 1][j] && maze[i][j + 1]) { code[j - 1] = code[j] = 13; hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]); } } }}void dpblock(int i, int j, int cur){ int k; for(k = 0; k < hm[cur].size; k ++) { decode(code, M, hm[cur].state[k]); code[j - 1] = code[j] = 0; if(j == M) shift(code, M); hm[cur ^ 1].push(encode(code, M), hm[cur].f[k]); }}void solve(){ int i, j, cur = 0; long long ans = 0; memset(code, 0, sizeof(code)); code[1] = code[M] = 1; hm[cur].init(); hm[cur].push(encode(code, M), 1); for(i = N; i >= 1; i --) for(j = 1; j <= M; j ++) { hm[cur ^ 1].init(); if(maze[i][j]) dpblank(i, j, cur); else dpblock(i, j, cur); cur ^= 1; } for(i = 0; i < hm[cur].size; i ++) ans += hm[cur].f[i]; printf("%lld\n", ans);}int main(){ for(;;) { scanf("%d%d", &N, &M); if(!N && !M) break; init(); solve(); } return 0;}