[HDOJ5929]Basic Data Structure（双向队列，规律）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5929

题意：维护一个栈，支持往栈里塞 0/1 ，弹栈顶，翻转栈，询问从栈底到栈顶按顺序 NAND 的值。

题解：只要知道最后的 00 后面 11 的个数的奇偶性就行。可以用链表把所有 00 的位置存下来。

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 
  4 typedef long long LL;
  5 
  6 const int maxn = 866666;
  7 int n;
  8 int q[maxn], l, r;
  9 int ll[maxn], rr[maxn];
 10 int zerol, zeror;
 11 char cmd[10];
 12 
 13 int main() {
 14   // freopen("in", "r", stdin);
 15   int T, _ = 1;
 16   scanf("%d", &T);
 17   while(T--) {
 18     scanf("%d", &n);
 19     l = maxn/2 + 1; r = maxn/2;
 20     int x, cur;
 21     zerol = zeror = -1;
 22     int dir = 0, zero = 0;
 23     printf("Case #%d:\n", _++);
 24     while(n--) {
 25       scanf("%s", cmd);
 26       if(strcmp(cmd, "PUSH")==0) {
 27         scanf("%d", &x);
 28         if(!dir) {
 29           q[--l] = x;
 30           cur = l;
 31         }
 32         else {
 33           q[++r] = x;
 34           cur = r;
 35         }
 36         if(x == 0) {
 37           zero++;
 38           if(zero == 1) {
 39             zerol = zeror = cur;
 40             ll[cur] = rr[cur] = -1;
 41           }
 42           else {
 43             if(!dir) {
 44               ll[zerol] = cur; rr[cur] = zerol;
 45               ll[cur] = -1; zerol = cur;
 46             }
 47             else {
 48               rr[zeror] = cur; ll[cur] = zeror;
 49               rr[cur] = -1; zeror = cur;
 50             }
 51           }
 52         }
 53       }
 54       if(strcmp(cmd, "POP")==0) {
 55         if(!dir) {
 56           x = q[l];
 57           cur = l++;
 58         }
 59         else {
 60           x = q[r];
 61           cur = r--;
 62         }
 63         if(x == 0) {
 64           zero--;
 65           if(zero == 0) zerol = zeror = -1;
 66           else {
 67             if(!dir) {
 68               ll[rr[cur]] = -1;
 69               zerol = rr[cur];
 70             }
 71             else {
 72               rr[ll[cur]] = -1;
 73               zeror = ll[cur];
 74             }
 75           }
 76         }
 77       }
 78       else if(strcmp(cmd, "REVERSE")==0) dir ^= 1;
 79       else if(strcmp(cmd, "QUERY")==0) {
 80         int siz = r - l + 1, one;
 81         if(!siz) printf("Invalid.\n");
 82         else {
 83           if(!zero) printf("%d\n", siz%2);
 84           else {
 85             int tmp;
 86             if(!dir) {
 87               one = r - zeror;
 88               tmp = (zeror == l);
 89             }
 90             else {
 91               one = zerol - l;
 92               tmp = (zerol == r);
 93             }
 94             printf("%d\n", one % 2 == tmp);
 95           }
 96         }
 97       }
 98     }
 99   }
100   return 0;
101 }

 

posted @ 2016-10-06 22:15  Kirai  阅读(293)  评论(0)  编辑  收藏  举报