本文主要是介绍[ural1519]Formula 1 插头DP，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
这里用的是最小整数表示法 具体方法参考cdq论文- -
if(U)里面多打了个else引发了一场腥风血雨啊QAQ
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define SF scanf
#define PF printf
#define bit(x) (1<<(x))
#define update dp[cur].push(encode() >> 3*(j==m) , dp[cur^1].f[k])
using namespace std;
typedef long long LL;
const int MAXNODE = 1000000;
const int MAXN = 12;
const int MOD = 233333;
struct Hash {int sz, adj[MOD+1], next[MAXNODE+10];LL state[MAXNODE+10], f[MAXNODE+10];void clr() { sz = 0; memset(adj, -1, sizeof(adj)); }int push(LL s, LL val) {int u = s % MOD;for(int i = adj[u]; ~i; i = next[i]) if(state[i] == s)return f[i] += val;state[sz] = s; f[sz] = val;next[sz] = adj[u]; adj[u] = sz++;return f[sz-1];}
} dp[2];
int n, m, M[MAXN+10][MAXN+10], ex, ey, cur;
int a[MAXN+10], num[MAXN+10];
void decode(LL s) {for(int i = m; i >= 0; i--) a[i] = s & 7, s >>= 3;
}
LL encode() {memset(num, 0, sizeof(num));int k = 0;LL ret = 0;for(int i = 0; i <= m; i++) {if(!num[a[i]] && a[i]) num[a[i]] = ++k;ret = (ret << 3) | num[a[i]];}return ret;
}
void change(int x, int y) {for(int i = 0; i <= m; i++) if(a[i] == x) a[i] = y;
}
void dpblank(int i, int j) {for(int k = 0; k < dp[cur^1].sz; k++) {decode(dp[cur^1].state[k]);int &L = a[j-1], &U = a[j];if(L && U) {if(L == U) {if(i == ex && j == ey) {L = U = 0;update;}}else {change(L, U); L = U = 0;update;}}else if(L) {if(M[i+1][j]) dp[cur].push(dp[cur^1].state[k] >> 3 * (j==m), dp[cur^1].f[k]);if(M[i][j+1]) {U = L; L = 0;update;}}else if(U) {if(M[i][j+1]) dp[cur].push(dp[cur^1].state[k] >> 3 * (j==m), dp[cur^1].f[k]);if(M[i+1][j]) {L = U; U = 0;update;}}else {if(M[i][j+1] && M[i+1][j]) {L = U = m+1;update;}}}
}
void dpblock(int i, int j) {for(int k = 0; k < dp[cur^1].sz; k++)dp[cur].push(dp[cur^1].state[k] >> (3 * (j == m)), dp[cur^1].f[k]);
}
int main() {char ch;SF("%d%d", &n, &m);for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++) {SF(" %c", &ch);if(ch == '.') M[ex = i][ey = j] = 1;else M[i][j] = 0;}dp[0].clr(); dp[0].push(0, 1);for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++) {cur ^= 1; dp[cur].clr();if(M[i][j]) dpblank(i, j);else dpblock(i, j);}LL ans = 0;for(int i = 0; i < dp[cur^1].sz; i++) ans += dp[cur^1].f[i];cout << ans;
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