CodeForces 149D Coloring Brackets

Coloring Brackets

题解：

dp[ l ] [ r ] [ lc ] [ rc ]  

代表的是第在区间[ l , r] 的情况下 左端点颜色是lc， 右端点颜色是rc的方案数是多少。

然后记忆化DP。

将一个序列拆成一个个匹配的序列。

为了防止一开始序列不匹配，所以从2个虚拟的地方开始计算。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1000;
char s[N];
int dp[N][N][3][3];
int link[N];
stack<int> sta;
int solve(int l, int r, int cl, int cr){
    if(~dp[l][r][cl][cr]) return dp[l][r][cl][cr];
    if(l+1 == r){
        dp[l][r][cl][cr] = 1;
        return 1;
    }
    int now = 0, pre = 1;
    LL t[2][3];
    for(int i = 0; i < 3; ++i)
        t[0][i] = t[1][i] = 0;
    t[0][cl] = 1;
    for(int ll = l + 1, rr = link[ll]; ll < r; ll = rr + 1, rr = link[ll]){
        swap(now, pre);
        for(int i = 0; i < 3; ++i)
            t[now][i] = 0;
        t[now][0] = t[pre][0] * solve(ll, rr, 1, 0) + t[pre][0] * solve(ll, rr, 2, 0)
                 + t[pre][1] * solve(ll, rr, 2, 0) + t[pre][2] * solve(ll, rr, 1, 0);
        t[now][0] %= mod;
        t[now][1] = t[pre][0] * solve(ll, rr, 0, 1) + t[pre][1] * solve(ll, rr, 0, 1) + t[pre][2] * solve(ll, rr, 0, 1);
        t[now][1] %= mod;
        t[now][2] = t[pre][0] * solve(ll, rr, 0, 2) + t[pre][1] * solve(ll, rr, 0, 2) + t[pre][2] * solve(ll, rr, 0, 2);
        t[now][2] %= mod;
    }
    if(cr == 0) dp[l][r][cl][cr] = (t[now][0] + t[now][1] + t[now][2]) % mod;
    else if(cr == 1) dp[l][r][cl][cr] = (t[now][0] + t[now][2]) % mod;
    else dp[l][r][cl][cr] = (t[now][0] + t[now][1]) % mod;
    return dp[l][r][cl][cr];
}
int main(){
    memset(dp, -1, sizeof dp);
    scanf("%s", s+1);
    int n = strlen(s+1);
    for(int i = 1; i <= n; ++i){
        if(s[i] == '(') sta.push(i);
        else {
            int j = sta.top();
            link[i] = j; link[j] = i;
            sta.pop();
        }
    }
    int ans = 1ll * solve(0, n+1, 0, 0);
    //int ans = (1ll * solve(1, n, 0, 1) + solve(1, n, 1, 0) + solve(1, n, 2, 0) + solve(1, n, 0, 2)) % mod;
    printf("%d
", ans);
    return 0;
}