【USACO 2015 Open Gold】Palindromic Paths 回文路径

【USACO 2015 Open Gold】Palindromic Paths 动态规划

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辗转山河弋流歌

题意：

从 n×n 的矩阵 左上角走到右下角会有一个长度 n+n+1 的字符串，问有多少种走法使得路径字符串为回文？

题解：

f(i,j,k,l) 表示起点横着走 i 步，竖着走 j 步，终点竖着走 k 步，横着走 l 步时的回文方案数。
然后跑动态规划时 f(i,j,k,l) 可以更新
f(i+1,j,k+1,l)、f(i+1,j,k,l+1)、f(i,j+1,k+1,l)、f(i,j+1,k,l+1)
跑动态规划的顺序是 (i+j) 从小到大。

然后时间复杂度是 O(n4)，但是我们发现(i+j==k+l) ，所以可以少枚举一个 l ，然后时空复杂度就变成 O(n3) 了，但是这道题64MB，会MLE，然后我们发现 (i+j) 满足滚动数组的性质，空间复杂度变成了 O(n2) 。

代码：

<code class="language-cpp hljs  has-numbering"><span class="hljs-preprocessor">#include <cstdio></span><span class="hljs-preprocessor">#include <cstring></span><span class="hljs-preprocessor">#include <iostream></span><span class="hljs-preprocessor">#include <algorithm></span><span class="hljs-preprocessor">#define N 505</span><span class="hljs-preprocessor">#define mod 1000000007</span><span class="hljs-keyword">using</span> <span class="hljs-keyword">namespace</span> <span class="hljs-built_in">std</span>;<span class="hljs-keyword">int</span> n,f[<span class="hljs-number">2</span>][N][N];<span class="hljs-keyword">char</span> s[N][N];<span class="hljs-preprocessor">#define add(a,b) (a=(a+b)%mod)</span><span class="hljs-keyword">int</span> main(){    <span class="hljs-keyword">int</span> i,j,k,l;    <span class="hljs-keyword">int</span> a,b,x,y;    <span class="hljs-built_in">scanf</span>(<span class="hljs-string">"%d"</span>,&n);    <span class="hljs-keyword">for</span>(i=<span class="hljs-number">1</span>;i<=n;i++)<span class="hljs-built_in">scanf</span>(<span class="hljs-string">"%s"</span>,s[i]+<span class="hljs-number">1</span>);    <span class="hljs-keyword">if</span>(s[<span class="hljs-number">1</span>][<span class="hljs-number">1</span>]!=s[n][n])    {        <span class="hljs-built_in">puts</span>(<span class="hljs-string">"0"</span>);        <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;    }    <span class="hljs-keyword">int</span> now=<span class="hljs-number">1</span>,last=<span class="hljs-number">0</span>;    f[now][<span class="hljs-number">0</span>][<span class="hljs-number">0</span>]=<span class="hljs-number">1</span>;    <span class="hljs-keyword">for</span>(<span class="hljs-keyword">int</span> h=<span class="hljs-number">0</span>;h<n;h++)    {        now^=<span class="hljs-number">1</span>,last^=<span class="hljs-number">1</span>;        <span class="hljs-built_in">memset</span>(f[now],<span class="hljs-number">0</span>,<span class="hljs-keyword">sizeof</span> f[now]);        <span class="hljs-keyword">for</span>(i=<span class="hljs-number">0</span>;i<=h;i++)        {            j=h-i,a=<span class="hljs-number">1</span>+i,b=<span class="hljs-number">1</span>+j;            <span class="hljs-keyword">for</span>(k=<span class="hljs-number">0</span>;k<=h;k++)            {                l=h-k,x=n-l,y=n-k;                <span class="hljs-keyword">if</span>(s[a+<span class="hljs-number">1</span>][b]==s[x-<span class="hljs-number">1</span>][y])add(f[now][i+<span class="hljs-number">1</span>][ k ],f[last][i][k]);                <span class="hljs-keyword">if</span>(s[a+<span class="hljs-number">1</span>][b]==s[x][y-<span class="hljs-number">1</span>])add(f[now][i+<span class="hljs-number">1</span>][k+<span class="hljs-number">1</span>],f[last][i][k]);                <span class="hljs-keyword">if</span>(s[a][b+<span class="hljs-number">1</span>]==s[x-<span class="hljs-number">1</span>][y])add(f[now][ i ][ k ],f[last][i][k]);                <span class="hljs-keyword">if</span>(s[a][b+<span class="hljs-number">1</span>]==s[x][y-<span class="hljs-number">1</span>])add(f[now][ i ][k+<span class="hljs-number">1</span>],f[last][i][k]);            }        }    }    <span class="hljs-keyword">int</span> ans=<span class="hljs-number">0</span>;    <span class="hljs-keyword">for</span>(i=<span class="hljs-number">0</span>;i<n;i++)add(ans,f[last][i][i]);    <span class="hljs-built_in">cout</span><<ans<<endl;    <span class="hljs-keyword">return</span> <span class="hljs-number">0</span>;}</code>