POJ 1068 Parencodings

POJ 1068 Parencodings

#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 41;
int lr[maxn],rl[maxn],r[maxn],ans[maxn];//记录每个括号左侧的右括号数，右侧的左括号数，右括号的位置

int main(){
	int cas,n,num,cnt;
	scanf("%d",&cas);
	while(cas--){
		cnt = 1;
		scanf("%d",&n);
		for(int i = 1;i<=n;i++){
			scanf("%d",&num);
			for(int j = cnt;j<=i+num;j++){
				lr[j] = i-1;
				rl[j] = j==i+num?n-j+i:n-1-j+i;
			}
			r[i] = i+num;
			cnt = i+num+1;
		}
		for(int i = 1;i<=n;i++){
			int reg = -1; //记录与第i个右括号匹配的左括号位置
			for(int j = r[i]-1;j>0;j--){
				if((rl[j]-rl[r[i]])==(lr[r[i]]-lr[j])){ //找到与每个右括号匹配的左括号
					reg = j;
					break;
				}
			}
			ans[i] = lr[r[i]] - lr[reg] + 1;
		}
		for(int i = 1;i<n;i++)
			printf("%d ",ans[i]);
		printf("%d\n",ans[n]);
	}
}