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

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

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

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] <=  && R[i] >= ;
    f[i][j] = false;
    for(int k = i; k <= j; k++) {
        if( * (k - i) +  < L[i] ||  * (k - i) +  > R[i]) continue;
        if(dp(i + , k) && dp(k + , 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 *  + ] = ')';
    print(i + , i + cnt, start + , path[i + ][i + cnt]);
    print(i + cnt + , j, start +  * cnt + , path[i + cnt + ][j]);
}

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

/*
*/