POJ 1739 Tony's Tour（插头DP）

Description

A square township has been divided up into n*m(n rows and m columns) square plots (1<=N,M<=8),some of them are blocked, others are unblocked. The Farm is located in the lower left plot and the Market is located in the lower right plot. Tony takes her tour of the township going from Farm to Market by walking through every unblocked plot exactly once. 
Write a program that will count how many unique tours Betsy can take in going from Farm to Market. 

Input

The input contains several test cases. The first line of each test case contain two integer numbers n,m, denoting the number of rows and columns of the farm. The following n lines each contains m characters, describe the farm. A '#' means a blocked square, a '.' means a unblocked square. 
The last test case is followed by two zeros. 

Output

For each test case output the answer on a single line.

 

题目大意：找到一条路径，经过所有非阻塞点，从右下到达左下。

思路：在最后加两排

.#######.

.......

就和 http://www.cnblogs.com/oyking/p/3394135.html一样了

 

代码（16MS）：

  1 #include <cstdio>

  2 #include <iostream>

  3 #include <algorithm>

  4 #include <cstring>

  5 using namespace std;

  6 typedef long long LL;

  7 

  8 const int MAXN = 12;

  9 const int SIZEH = 13131;

 10 const int MAXH = 20010;

 11 

 12 struct hash_map {

 13     int head[SIZEH], size;

 14     int state[MAXH], next[MAXH];

 15     LL val[MAXH];

 16 

 17     void init() {

 18         memset(head, -1, sizeof(head));

 19         size = 0;

 20     }

 21 

 22     void insert(int st, LL sv) {

 23         int h = st % SIZEH;

 24         for(int p = head[h]; ~p; p = next[p]) {

 25             if(state[p] == st) {

 26                 val[p] += sv;

 27                 return ;

 28             }

 29         }

 30         state[size] = st; val[size] = sv; next[size] = head[h]; head[h] = size++;

 31     }

 32 } hashmap[2];

 33 

 34 hash_map *cur, *last;

 35 int acc[] = {0, -1, 1, 0};

 36 char mat[MAXN][MAXN];

 37 int n, m, en, em;

 38 

 39 int getB(int state, int i) {

 40     return (state >> (i << 1)) & 3;

 41 }

 42 

 43 void setB(int &state, int i, int val) {

 44     state = (state & ~(3 << (i << 1))) | (val << (i << 1));

 45 }

 46 

 47 int getLB(int state, int i) {

 48     int ret = i, cnt = 1;

 49     while(cnt) {

 50         --ret;

 51         cnt += acc[getB(state, ret)];

 52     }

 53     return ret;

 54 }

 55 

 56 int getRB(int state, int i) {

 57     int ret = i, cnt = -1;

 58     while(cnt) {

 59         ++ret;

 60         cnt += acc[getB(state, ret)];

 61     }

 62     return ret;

 63 }

 64 

 65 void update(int x, int y, int state, LL tv) {

 66     int left = getB(state, y);

 67     int up = getB(state, y + 1);

 68     if(mat[x][y] == '#') {

 69         if(left == 0 && up == 0) cur->insert(state, tv);

 70         return ;

 71     }

 72     if(left == 0 && up == 0) {

 73         if(x == n - 1 || y == m - 1) return ;

 74         int newState = state;

 75         setB(newState, y, 1);

 76         setB(newState, y + 1, 2);

 77         cur->insert(newState, tv);

 78     } else if(left == 0 || up == 0) {

 79         if(x < n - 1) {

 80             int newState = state;

 81             setB(newState, y, up + left);

 82             setB(newState, y + 1, 0);

 83             cur->insert(newState, tv);

 84         }

 85         if(y < m - 1) {

 86             int newState = state;

 87             setB(newState, y, 0);

 88             setB(newState, y + 1, up + left);

 89             cur->insert(newState, tv);

 90         }

 91     } else {

 92         int newState = state;

 93         setB(newState, y, 0);

 94         setB(newState, y + 1, 0);

 95         if(left == 1 && up == 1) setB(newState, getRB(state, y + 1), 1);

 96         if(left == 1 && up == 2 && !(x == en && y == em)) return ;

 97         if(left == 2 && up == 2) setB(newState, getLB(state, y), 2);

 98         cur->insert(newState, tv);

 99     }

100 }

101 

102 void findend() {

103     for(en = n - 1; en >= 0; --en)

104         for(em = m - 1; em >= 0; --em) if(mat[en][em] != '#') return ;

105 }

106 

107 LL solve() {

108     findend();

109     cur = hashmap, last = hashmap + 1;

110     last->init();

111     last->insert(0, 1);

112     for(int i = 0; i < n; ++i) {

113         int sz = last->size;

114         for(int k = 0; k < sz; ++k) last->state[k] <<= 2;

115         for(int j = 0; j < m; ++j) {

116             cur->init();

117             sz = last->size;

118             for(int k = 0; k < sz; ++k)

119                 update(i, j, last->state[k], last->val[k]);

120             swap(cur, last);

121         }

122     }

123     for(int k = 0; k < last->size; ++k)

124         if(last->state[k] == 0) return last->val[k];

125     return 0;

126 }

127 

128 int main() {

129     while(scanf("%d%d", &n, &m) != EOF) {

130         if(n == 0 && m == 0) break;

131         memset(mat, 0, sizeof(mat));

132         for(int i = 0; i < n; ++i) scanf("%s", mat[i]);

133         for(int i = 1; i < m - 1; ++i) mat[n][i] = '#';

134         n += 2;

135         cout<<solve()<<endl;

136     }

137 }

View Code