11234UVa表达式——动态 &RE原因

这题的数据量应该很大，之前char ans[10000]；

char str[10000];

Node* stack[1000];   //10000      AC

Node* queue[1000]; //10000       AC

一直runtime error!

AC后用时：0.900

相关参考：动态STL  http://blog.youkuaiyun.com/ultimater/article/details/8015040       用时：0.08

静态 http://blog.youkuaiyun.com/u011345461/article/details/10063707         用时：0.84

#include<stdio.h>
#include<ctype.h>
#include<string.h>
struct Node
{
	Node *left,*right;
	char ch;
	Node(char c, Node* l = 0, Node* r = 0)
	{
		ch = c;
		left = l;
		right = r;
	}
};
char ans[10000];
char str[10000];
Node* stack[10000];
Node* queue[10000];
void cleanup(Node* h);
int main()
{
	//
	freopen("input.txt","r",stdin);
	int numtest;
	scanf("%d", &numtest);
	Node* first,* sec;
	for(int i = 0; i < numtest; i++)
	{
		scanf("%s", str);
		int top = 0;
		for(int j = 0; j < strlen(str); j++)
		{
			if(islower(str[j]) )
				stack[++top] = new Node(str[j]);
			else if(isupper(str[j]) )
			{
				first = stack[top--];
				sec = stack[top--];
				stack[++top] = new Node(str[j], sec, first);
			}
		}
		int f = 0;
		int r = 1;
		memset(queue, 0, sizeof(queue) );
		queue[f] = stack[1];
		int num = 0;
		while(f < r)
		{
			Node* temp = queue[f++];
			if(temp->left) queue[r++] = temp->left;
			if(temp->right) queue[r++] = temp->right;
			ans[num++] = temp->ch;
		}
		for(int  k = num -1; k >= 0; k--)
		{
			putchar(ans[k]);
		}
		putchar('\n');
		cleanup(stack[1]);
	}
}
void cleanup(Node* h)
{
	if(!h) return;
	cleanup(h->left);
	cleanup(h->right);
	delete h;
}