通道：http://poj.org/problem?id=1739
题意：左下角走到右下角路径数，单回路。做法就是我们新添2行后寻找回路就可以啦 。
代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4
5 using namespace std;
6
7 const int MAX_N = 13;
8 const int MAX_M = 13;
9 const int HASH = 10007;
10 const int MAX_S = 1000007;
11
12 struct node {
13 int head[HASH], nxt[MAX_S];
14 long long dp[MAX_S], st[MAX_S];
15 int cnt;
16 void init() {
17 memset(head, -1, sizeof head);
18 cnt = 0;
19 }
20 void push(long long s, long long v) {
21 int now = s % HASH;
22 for(int i = head[now]; ~i; i = nxt[i]) if(st[i] == s) {
23 dp[i] += v;
24 return ;
25 }
26 st[cnt] = s; dp[cnt] = v;
27 nxt[cnt] = head[now];
28 head[now] = cnt++;
29 }
30 }d[2];
31
32 int n, m;
33 int ex, ey;
34
35 int find_pos(long long s, int p) {
36 return (s >> (p << 1)) & 3;
37 }
38
39 void tp(long long &s, int p, long long v) {
40 s &= (~(3ll << (p << 1)));
41 s |= (v << (p << 1));
42 }
43
44 int find_r(long long s, int p) {
45 int cnt = 0;
46 for(int i = p; i <= m; ++i) {
47 if(find_pos(s, i) == 1) ++cnt;
48 else if(find_pos(s, i) == 2) --cnt;
49 if(!cnt) return i;
50 }
51 }
52
53 int find_l(long long s, int p) {
54 int cnt = 0;
55 for(int i = p; i >= 0; --i) {
56 if(find_pos(s, i) == 2) ++cnt;
57 else if(find_pos(s, i) == 1) --cnt;
58 if(!cnt) return i;
59 }
60 }
61
62 void blank(int i, int j, int cur) {
63 for(int k = 0; k < d[cur].cnt; ++k) {
64 long long t = d[cur].st[k];
65 int l = find_pos(t, j - 1), r = find_pos(t, j);
66 if(l && r) {
67 if(l == 1 && r == 1) {
68 int tpos = find_r(t, j);
69 tp(t, j - 1, 0); tp(t, j, 0); tp(t, tpos, 1);
70 d[cur ^ 1].push(t, d[cur].dp[k]);
71 } else if(l == 2 && r == 1) {
72 tp(t, j - 1, 0); tp(t, j, 0);
73 d[cur ^ 1].push(t, d[cur].dp[k]);
74 } else if(l == 2 && r == 2) {
75 int tpos = find_l(t, j - 1);
76 tp(t, j - 1, 0); tp(t, j, 0); tp(t, tpos, 2);
77 d[cur ^ 1].push(t, d[cur].dp[k]);
78 } else { // 最后一个非障碍格子
79 tp(t, j - 1, 0); tp(t, j, 0);
80 // 2位为1代表有1个回路，4位为1代表有2个回路
81 if (!t) if (i == ex && j == ey) d[cur ^ 1].push(t, d[cur].dp[k]);
82 }
83 } else if(l) {
84 if(i < n) {
85 d[cur ^ 1].push(t, d[cur].dp[k]);
86 }
87 if(j < m) {
88 tp(t, j - 1, 0); tp(t, j, l);
89 d[cur ^ 1].push(t, d[cur].dp[k]);
90 }
91 } else if(r) {
92 if(j < m) {
93 d[cur ^ 1].push(t, d[cur].dp[k]);
94 }
95 if(i < n) {
96 tp(t, j - 1, r); tp(t, j, 0);
97 d[cur ^ 1].push(t, d[cur].dp[k]);
98 }
99 } else { // 新建
100 if(i < n && j < m) {
101 tp(t, j - 1, 1); tp(t, j, 2);
102 d[cur ^ 1].push(t, d[cur].dp[k]);
103 }
104 }
105 }
106 }
107
108 void block(int i, int j, int cur) {
109 for (int k = 0; k < d[cur].cnt; ++k) {
110 long long t = d[cur].st[k];
111 int l = find_pos(t, j - 1), r = find_pos(t, j);
112 if (!l && !r) d[cur ^ 1].push(t, d[cur].dp[k]);
113 }
114 }
115
116 char str[17];
117 int a[MAX_N][MAX_M];
118
119 int main() {
120 while (2 == scanf("%d%d", &n, &m)) {
121 if (0 == m && 0 == m) break;
122 memset(a, 0, sizeof a);
123 for (int i = 1; i <= n; ++i) {
124 scanf("%s", str + 1);
125 for (int j = 1; j <= m; ++j)
126 a[i][j] = str[j] == ‘.‘;
127 }
128 for (int i = 1; i <= m; ++i)
129 a[n + 2][i] = 1;
130 for (int i = 2; i < m; ++i)
131 a[n + 1][i] = 0;
132 a[n + 1][1] = a[n + 1][m] = 1;
133 int cur = 0;
134 long long ans = 0;
135 d[cur].init();
136 d[cur].push(0, 1);
137 n = n + 2;
138 ex = n, ey = m;
139 for (int i = 1; i <= n; ++i) {
140 for (int j = 1; j <= m; ++j) {
141 d[cur ^ 1].init();
142 if (a[i][j]) blank(i, j, cur);
143 else block(i, j, cur);
144 cur ^= 1;
145 }
146 for (int j = 0; j < d[cur].cnt; ++j)
147 d[cur].st[j] <<= 2;
148 }
149 for(int i = 0; i < d[cur].cnt; ++i)
150 ans += d[cur].dp[i];
151 printf("%I64d\n", ans);
152 }
153 return 0;
154 }
【插头DP】 POJ 1739 Tony's Tour
时间： 2024-11-05 18:39:09