Codeforces 570E Pig and palindromes DP

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Test instructions

The letter matrix for the given n*m.

The path from the upper-left corner to the lower-right corner is a palindrome.

Ideas:

Obviously to the head of the DP at the same time, the path 1 is from the upper left to line J, Path 2 is from the lower right corner to the K line

Dp[i][j][k] indicates that the path length is I, path 1 from the upper left to the first row J, path 2 from the lower right to the K line, and Paths 1 and 2 are the number of matching methods.

For paths 1, 2, you want to split the parity when merging.

#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include <stdio.h> #include <iostream> #include <algorithm> #include <sstream> #include <stdlib.h> #include <string.h> #include &LT;LIMITS.H&G  T #include <vector> #include <string> #include <time.h> #include <math.h> #include &LT;IOMANIP&G  T #include <queue> #include <stack> #include <set> #include <map> const int inf = 1e9;const Doub Le EPS = 1e-8;const double pi = ACOs ( -1.0); template <class t>inline BOOL Rd (T &ret) {char c; int sgn;if (c = Get char (), c = = EOF) return 0;while (c! = '-' && (c< ' 0 ' | | c> ' 9 ')) C = GetChar (); sgn = (c = = '-')? -1:1;ret = (c = = '-')? 0: (C-' 0 '); while (c = GetChar (), C >= ' 0 ' &&c <= ' 9 ') ret = ret * + (C-' 0 '); ret *= Sgn;return 1;} Template <class t>inline void pt (T x) {if (x < 0) {Putchar ('-'); x =-X;} if (x > 9) pt (X/10);p Utchar (x% 10 + ' 0 ');}using namespace Std;const int N = 520;const int mod = 1e9 + 7;typedef long long ll;typedef pair<int, int> pii;void A DD (int &x, int y) {x + = y;if (x >= mod) x-= mod;} int mul (int x, int y) {x = (ll) X*y%mod;return X;} int N, M;char s[n][n];vector<pii>g[1005];int step[4][2] = {0, 1, 1, 0, 0,-1,-1, 0}, Dis[n][n];bool inmap (int x, int y) {return 1 <= x&&x <= n && 1 <= y&&y <= m;} void BFs () {queue<int>qx, Qy;qx.push (1), Qy.push (1);d is[1][1] = 1;while (!qx.empty ()) {int UX = Qx.front (), Uy = qy. Front (); G[dis[ux][uy]].push_back ({UX, uy}); Qx.pop (); Qy.pop (); for (int i = 0; i < 2; i++) {int VX = UX + step[i][0], vy = Uy + step[i][1];if (INMAP (VX, VY) &&!dis[ Vx][vy]) {Dis[vx][vy] = Dis[ux][uy] + 1;qx.push (VX); Qy.push (VY);}}} int Dp[2][n][n];int Main () {rd (N); Rd (M); for (int i = 1; I <= N; i++) scanf ("%s", S[i] + 1), if (s[1][1]! = s[n][m]) {p UTS ("0"); return 0; }S[0][1] = ' a '; S[1][0] = ' Z ' + 1;s[n + 1][m] = ' a '; S[n][m + 1] = ' Z ' + 1;BFS (); int cur = 0, old = 1;memset (Dp[cur], 0, sizeof dp[cur]);DP [cur][1][n] = 1;int L = 1, R = n + M- 1;for (int i = 2; I <= (n + m)/2; i++, l++, r--) {swap (cur, old), memset (Dp[cur], 0, sizeof dp[cur]); for (auto U:g[l] {int x = u.first, y = u.second;for (auto v:g[r]) {int tx = V.first, Ty = v.second;if (dp[old][x][tx] = = 0) continue;for ( int a = 0; A < 2;  a++) {int UX = x + step[a][0], Uy = y + step[a][1];if (!inmap (UX, UY)) continue;for (int b = 2; b < 4; b++) {int VX = tx + Step[b][0], vy = ty + step[b][1];if (!INMAP (VX, VY)) continue;if (S[ux][uy]! = S[vx][vy]) continue;add (DP[CUR][UX][VX], dp[ OLD][X][TX]);}}}} int ans = 0;if ((n + m) & 1) {for (int i = 1; I <= n; i++) Add (ans, dp[cur][i][i]); for (int i = 1; I <= n; i++) add (ans, dp[cur][i][i + 1]);} else {for (int i = 1; I <= n; i++) Add (ans, dp[cur][i][i]);} PT (ANS); return 0;}

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Codeforces 570E Pig and palindromes DP