CF149D Coloring Brackets

DP神题。。。
设(dp[i][j][0/1/2][0/1/2])表示([i,j])这个区间内端点取不染色/染红色/染蓝色三个状态然后转移。一个新(trick)就是这里的转移只要考虑这个区间内的串是合法的就可以了

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

#define R register
#define LL long long
const int inf = 0x3f3f3f3f;
const int N = 700 + 5;
const int P = 1e9 + 7;

inline int read() {
	char a = getchar(); int x = 0,f = 1;
	for(; a > '9' || a < '0'; a = getchar()) if(a == '-') f = -1;
	for(; a >= '0' && a <= '9' ;a = getchar()) x = x * 10 + a - '0';
	return x * f;
}

int n;
char s[N];
int f[N][N][3][3]; // 700 * 700 * 3 * 3 
int nxt[N];
int st[N], top;

inline void dp(int l, int r) {
	if(l + 1 == r) {
		f[l][r][0][1] = f[l][r][1][0] = f[l][r][2][0] = f[l][r][0][2] = 1;
		return ;
	}
	if(nxt[l] == r) {
		dp(l + 1, r - 1);
		for(R int i = 0; i < 3; i++) 
			for(R int j = 0; j < 3; j++) {
				if(j != 1) f[l][r][0][1] = (f[l][r][0][1] + f[l + 1][r - 1][i][j]) % P;
				if(i != 1) f[l][r][1][0] = (f[l][r][1][0] + f[l + 1][r - 1][i][j]) % P;
				if(j != 2) f[l][r][0][2] = (f[l][r][0][2] + f[l + 1][r - 1][i][j]) % P;
				if(i != 2) f[l][r][2][0] = (f[l][r][2][0] + f[l + 1][r - 1][i][j]) % P;
			}
		return ;
	}
	else {
		int m = nxt[l];
		dp(l, m);
		dp(m + 1, r);
		for(R int i = 0; i < 3; i++)
			for(R int j = 0; j < 3; j++)
				for(R int u = 0; u < 3; u++)
					for(R int v = 0 ; v < 3 ; v++ ) {
						if(u == v && u != 0) continue;
						f[l][r][i][j] = (f[l][r][i][j] + (1LL * f[l][m][i][u] * f[m + 1][r][v][j]) % P ) % P;
					}
	}
}

int main() {
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	scanf("%s", s + 1);
	n = strlen(s + 1);
	for(R int i = 1; i <= n; i++) {
		if(s[i] == '(') st[++top] = i;
		else {
			nxt[st[top]] = i;
			top--;
		}
	}
	dp(1, n);
	int ans = 0;
	for(R int i = 0; i < 3; i++ )
		for(R int j = 0; j < 3; j++)
			ans = ( ans + f[1][n][i][j] ) % P;
	printf("%d
", ans);
	return 0;	
}