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
.#######.
.......
 
代码(16MS):
 #include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL; const int MAXN = ;
const int SIZEH = ;
const int MAXH = ; struct hash_map {
int head[SIZEH], size;
int state[MAXH], next[MAXH];
LL val[MAXH]; void init() {
memset(head, -, sizeof(head));
size = ;
} void insert(int st, LL sv) {
int h = st % SIZEH;
for(int p = head[h]; ~p; p = next[p]) {
if(state[p] == st) {
val[p] += sv;
return ;
}
}
state[size] = st; val[size] = sv; next[size] = head[h]; head[h] = size++;
}
} hashmap[]; hash_map *cur, *last;
int acc[] = {, -, , };
char mat[MAXN][MAXN];
int n, m, en, em; int getB(int state, int i) {
return (state >> (i << )) & ;
} void setB(int &state, int i, int val) {
state = (state & ~( << (i << ))) | (val << (i << ));
} int getLB(int state, int i) {
int ret = i, cnt = ;
while(cnt) {
--ret;
cnt += acc[getB(state, ret)];
}
return ret;
} int getRB(int state, int i) {
int ret = i, cnt = -;
while(cnt) {
++ret;
cnt += acc[getB(state, ret)];
}
return ret;
} void update(int x, int y, int state, LL tv) {
int left = getB(state, y);
int up = getB(state, y + );
if(mat[x][y] == '#') {
if(left == && up == ) cur->insert(state, tv);
return ;
}
if(left == && up == ) {
if(x == n - || y == m - ) return ;
int newState = state;
setB(newState, y, );
setB(newState, y + , );
cur->insert(newState, tv);
} else if(left == || up == ) {
if(x < n - ) {
int newState = state;
setB(newState, y, up + left);
setB(newState, y + , );
cur->insert(newState, tv);
}
if(y < m - ) {
int newState = state;
setB(newState, y, );
setB(newState, y + , up + left);
cur->insert(newState, tv);
}
} else {
int newState = state;
setB(newState, y, );
setB(newState, y + , );
if(left == && up == ) setB(newState, getRB(state, y + ), );
if(left == && up == && !(x == en && y == em)) return ;
if(left == && up == ) setB(newState, getLB(state, y), );
cur->insert(newState, tv);
}
} void findend() {
for(en = n - ; en >= ; --en)
for(em = m - ; em >= ; --em) if(mat[en][em] != '#') return ;
} LL solve() {
findend();
cur = hashmap, last = hashmap + ;
last->init();
last->insert(, );
for(int i = ; i < n; ++i) {
int sz = last->size;
for(int k = ; k < sz; ++k) last->state[k] <<= ;
for(int j = ; j < m; ++j) {
cur->init();
sz = last->size;
for(int k = ; k < sz; ++k)
update(i, j, last->state[k], last->val[k]);
swap(cur, last);
}
}
for(int k = ; k < last->size; ++k)
if(last->state[k] == ) return last->val[k];
return ;
} int main() {
while(scanf("%d%d", &n, &m) != EOF) {
if(n == && m == ) break;
memset(mat, , sizeof(mat));
for(int i = ; i < n; ++i) scanf("%s", mat[i]);
for(int i = ; i < m - ; ++i) mat[n][i] = '#';
n += ;
cout<<solve()<<endl;
}
}

POJ 1739 Tony's Tour(插头DP)的更多相关文章

  1. POJ 1739 Tony's Tour (插头DP,轮廓线DP)

    题意:给一个n*m的矩阵,其中#是障碍格子,其他则是必走的格子,问从左下角的格子走到右下角的格子有多少种方式. 思路: 注意有可能答案是0,就是障碍格子阻挡住了去路. 插头DP有两种比较常见的表示连通 ...

  2. [POJ 1739] Tony's Tour

    Link: POJ 1739 传送门 Solution: 这题除了一开始的预处理,基本上就是插头$dp$的模板题了 由于插头$dp$求的是$Hamilton$回路,而此题有起点和终点的限制 于是可以构 ...

  3. POJ 1739 Tony's Tour (DP)

    题意:从左下角到右下角有多少种走法. 析:特殊处理左下角和右下角即可. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000 ...

  4. POJ 2411 Mondriaan's Dream 插头dp

    题目链接: http://poj.org/problem?id=2411 Mondriaan's Dream Time Limit: 3000MSMemory Limit: 65536K 问题描述 S ...

  5. POJ 3133 Manhattan Wiring (插头DP,轮廓线,经典)

    题意:给一个n*m的矩阵,每个格子中有1个数,可能是0或2或3,出现2的格子数为2个,出现3的格子数为2个,要求将两个2相连,两个3相连,求不交叉的最短路(起终点只算0.5长,其他算1). 思路: 这 ...

  6. 【POJ】1739 Tony's Tour

    http://poj.org/problem?id=1739 题意:n×m的棋盘,'#'是障碍,'.'是空白,求左下角走到右下角且走过所有空白格子的方案数.(n,m<=8) #include & ...

  7. 插头DP专题

    建议入门的人先看cd琦的<基于连通性状态压缩的动态规划问题>.事半功倍. 插头DP其实是比较久以前听说的一个东西,当初是水了几道水题,最近打算温习一下,顺便看下能否入门之类. 插头DP建议 ...

  8. 插头DP题目泛做(为了对应WYD的课件)

    题目1:BZOJ 1814 URAL 1519 Formula 1 题目大意:给定一个N*M的棋盘,上面有障碍格子.求一个经过所有非障碍格子形成的回路的数量. 插头DP入门题.记录连通分量. #inc ...

  9. 【POJ】【1739】Tony's Tour

    插头DP 楼教主男人八题之一! 要求从左下角走到右下角的哈密顿路径数量. 啊嘞,我只会求哈密顿回路啊……这可怎么搞…… 容易想到:要是把起点和重点直接连上就变成一条回路了……那么我们就连一下~ 我们可 ...

随机推荐

  1. #leetcode刷题之路12-整数转罗马数字

    罗马数字包含以下七种字符: I, V, X, L,C,D 和 M. 字符 数值I 1V 5X 10L 50C 100D 500M 1000 例如, 罗马数字 2 写做 II ,即为两个并列的 1.12 ...

  2. hdu_4135_Co-prime

    Given a number N, you are asked to count the number of integers between A and B inclusive which are ...

  3. 复习宝典之SpringMVC

    查看更多宝典,请点击<金三银四,你的专属面试宝典> 第七章:SpringMVC MVC全名是Model View Controller,是模型(model)-视图(view)-控制器(co ...

  4. Ansible实现主备模式的高可用(Keepalived)

    前言 Ansible是一款极其简单的IT自动化运维工具,基于Python开发,集合了众多运维工具(puppet.cfengine.chef.func.fabric)的优点,实现了批量系统配置.批量程序 ...

  5. Pagination

    using System.Collections.Generic; namespace Oyang.Tool { public interface IPagination { int PageInde ...

  6. CSS中的动画

    1.transition 在CSS3中,可以通过transition为元素从一种样式变换为另外一种样式的过程添加效果. transition为简写属性,用于在一个属性中设置四个过渡属性,分别是: tr ...

  7. Git简单配置ssh秘钥

    执行以下命令: git config --global user.name "demo" git config --global user.email "demo@dem ...

  8. day 24 内置模块re

    1.正则表达式,匹配字符串 正则表达式是对字符串操作的一种逻辑公式.我们一般使用正则表达式对字符串镜子那个匹配和过滤,使用正则的优缺点: 优点: 灵活,功能性强,逻辑性强 缺点: 上手难.一旦上手,会 ...

  9. python--模块之sys与python解释器交互模块

    作用:sys模块是与python解释器交互的一个接口.它提供了一系列有关python运行环境的变量和函数. 常用函数:import sys sys.argv #命令行参数list,第一个元素是程序本身 ...

  10. BIOS简单讲解

    学习链接: http://www.xuetangx.com/courses/course-v1:TsinghuaX+30240243X+sp/courseware/1d95cdf6f0e9434488 ...