题目大意
有一个 n 行 m 列的格点图，你需要给每个点上染上 k 种颜色中的一种，要求没有两个相邻点颜色相同。给定第一行与最后一行的染色，试求总染色方案数。
思路
暴力预处理状态暴力转移可以得到80分的高分
这个时候司来了一句：
不要按行转移，按块转移就A了
于是改改改写了一个轮廓线
就把轮廓线上的颜色信息记录下来就好了，随便转移一下
注意终止状态也需要考虑轮廓线上的情况
需要把最后一个位置的轮廓加上
//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
typedef pair<int, int> pi;
typedef long long ll;
typedef double db;
#define fi first
#define se second
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
template <typename T>
void Read(T &x) {
bool w = 1;x = 0;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-') w = 0, c = getchar();
while (isdigit(c)) {
x = (x<<1) + (x<<3) + c -'0';
c = getchar();
}
if (!w) x = -x;
}
template <typename T>
void Write(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) Write(x / 10);
putchar(x % 10 + '0');
}
//----------------------------------------------
const int Mod = 376544743;
const int N = 1e2 + 10;
const int M = 10;
const int K = 1 << (M << 1);
int n, m, k;
int s[(int)1e5 + 10][2];
int add(int a, int b) {
return (a += b) >= Mod ? a - Mod : a;
}
int mul(int a, int b) {
return 1ll * a * b % Mod;
}
namespace Solve1 {
void solve() {
fu(i, 1, m) Read(s[i][0]);
fu(i, 1, m) Read(s[i][1]);
fu(i, 2, m) {
if (s[i][0] == s[i - 1][0] || s[i][1] == s[i - 1][1]) {
printf("0");
return;
}
}
fu(i, 1, m) {
if (s[i][0] != s[i][1]) {
if (n & 1) {
printf("0");
return;
}
} else {
if (!(n & 1)) {
printf("0");
return;
}
}
}
printf("1");
}
}
namespace Solve2 {
vector<int> group;
int dp[N][K];
bool trans[8800][8800];
void dfs(int tmp, int s) {
if (!tmp) {
group.push_back(s);
return;
}
fu(i, 0, k - 1)
if ((s & 3) != i) dfs(tmp - 1, (s << 2) | i);
}
bool check(int s1, int s2) {
fu(i, 1, m) {
if (!((s1 & 3) ^ (s2 & 3))) return 0;
s1 >>= 2;
s2 >>= 2;
}
return 1;
}
void solve() {
fu(i, 1, m) Read(s[i][0]);
fu(i, 1, m) Read(s[i][1]);
int bg = 0, ed = 0;
fu(i, 2, m) {
if (s[i][0] == s[i - 1][0] || s[i][1] == s[i - 1][1]) {
printf("0");
return;
}
}
fu(i, 1, m) {
bg = (bg << 2) + s[i][0];
ed = (ed << 2) + s[i][1];
}
fu(i, 0, k - 1) dfs(m - 1, i);
fv(i, group)
fu(j, 0, i - 1)
if (check(group[i], group[j]))
trans[i][j] = trans[j][i] = 1;
int posbg = 0, posed = 0;
fv(i, group) {
if (group[i] == bg) posbg = i;
if (group[i] == ed) posed = i;
}
dp[1][posbg] = 1;
fu(i, 2, (n >> 1)) {
fv(j, group) if (dp[i - 1][j]) {
fv(k, group) if (trans[j][k]) {
dp[i][k] = add(dp[i][k], dp[i - 1][j]);
}
}
}
dp[n][posed] = 1;
fd(i, n - 1, ((n >> 1) + 1)) {
fv(j, group) if (dp[i + 1][j]) {
fv(k, group) if (trans[j][k]) {
dp[i][k] = add(dp[i][k], dp[i + 1][j]);
}
}
}
int ans = 0;
fv(i, group) if (dp[n >> 1][i])
fv(j, group) if (trans[i][j])
ans = add(ans, mul(dp[n >> 1][i], dp[(n >> 1) + 1][j]));
Write(ans);
}
}
namespace Solve3 {
int dp[2][M][K];
void solve() {
fu(i, 1, m) Read(s[i][0]);
fu(i, 1, m) Read(s[i][1]);
int bg = 0, ed = 0, up = (1 << ((m + 1) << 1)) - 1;
fu(i, 2, m) {
if (s[i][0] == s[i - 1][0] || s[i][1] == s[i - 1][1]) {
printf("0");
return;
}
}
fd(i, m, 1) {
bg = (bg << 2) + s[i][0];
ed = (ed << 2) + s[i][1];
}
int ind = 0;
dp[ind][m][bg] = 1;
fu(i, 2, n) {
ind ^= 1;
memset(dp[ind], 0, sizeof(dp[ind]));
fu(s, 0, up) if (dp[ind ^ 1][m][s])
dp[ind][0][(s << 2) & up] = add(dp[ind][0][(s << 2) & up], dp[ind ^ 1][m][s]);
fu(j, 1, m) {
fu(s, 0, up) if (dp[ind][j - 1][s]) {
fu(p, 0, k - 1) if ((((j == 1) || (((s >> ((j - 1) << 1))) & 3) != p) && ((s >> (j << 1)) & 3) != p)) {
int nxt = (s & (up ^ (15 << ((j - 1) << 1)))) | ((p | (p << 2)) << ((j - 1) << 1));
dp[ind][j][nxt] = add(dp[ind][j][nxt], dp[ind][j - 1][s]);
}
}
}
}
Write(dp[ind][m][ed | (s[m][1] << (m << 1))]); // 结束状态不是ed 需要考虑最后的轮廓线状态
}
}
int main() {
#ifdef dream_maker
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
Read(n), Read(m), Read(k);
if (k == 2) Solve1::solve();
else if (m <= 0) Solve2::solve();
else Solve3::solve();
return 0;
}