Codeforces 508E Arthur and Brackets 区间dp

Arthur and Brackets

区间dp， dp[ i ][ j ]表示第 i 个括号到第 j 个括号之间的所有括号能不能形成一个合法方案。

然后dp就完事了。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using namespace std;

const int N = 600 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;

int f[N][N];
int path[N][N];

int n, L[N], R[N];
char ans[N << 1];

int dp(int i, int j) {
    if(i > j) return true;
    if(~f[i][j]) return f[i][j];
    if(i == j) return L[i] <= 1 && R[i] >= 1;
    f[i][j] = false;
    for(int k = i; k <= j; k++) {
        if(2 * (k - i) + 1 < L[i] || 2 * (k - i) + 1 > R[i]) continue;
        if(dp(i + 1, k) && dp(k + 1, j)) {
            f[i][j] = true;
            path[i][j] = k - i;
            break;
        }
    }
    return f[i][j];
}

void print(int i, int j, int start, int cnt) {
    if(i > j) return;
    ans[start] = '(';
    ans[start + cnt * 2 + 1] = ')';
    print(i + 1, i + cnt, start + 1, path[i + 1][i + cnt]);
    print(i + cnt + 1, j, start + 2 * cnt + 2, path[i + cnt + 1][j]);
}

int main() {
    memset(f, -1, sizeof(f));
    scanf("%d", &n);
    ans[2 * n + 1] = '\0';
    for(int i = 1; i <= n; i++) scanf("%d%d", &L[i], &R[i]);
    if(!dp(1, n)) puts("IMPOSSIBLE");
    else {
        print(1, n, 1, path[1][n]);
        puts(ans + 1);
    }
    return 0;
}

/*
*/