HDU 4699 Editor （2013多校10,1004题）

Editor

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 118    Accepted Submission(s): 38

Problem Description

 

Sample Input

8
I 2
I -1
I 1
Q 3
L
D
R
Q 2

 

Sample Output

2
3

Hint

The following diagram shows the status of sequence after each instruction:

 

Source

2013 Multi-University Training Contest 10

 

Recommend

zhuyuanchen520

 

这题只要用双向链表模拟一下。

查询最大前缀和，相当于维护一个单调队列。

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013/8/22 13:38:35
 File Name     :F:\2013ACM练习\2013多校10\1004.cpp
 ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;
 const int MAXN = ;

 int a[MAXN];
 int next[MAXN],pre[MAXN],tot;
 int NewNode()
 {
     next[tot] = -;
     pre[tot] = -;
     tot++;
     return tot-;
 }
 int root;
 int sum[MAXN];
 vector<int>vec;
 int now;
 int n;
 void init()
 {
     tot = ;
     root = NewNode();
     vec.clear();
     n = now = ;
 }

 void I(int x)
 {
     n++;
     int t = NewNode();
     a[t] = x;
     pre[t] = now;
     next[t] = next[now];
     if(next[now] != -)pre[next[now]] = t;
     next[now] = t;
     now = t;
     if(n == )
     {
         sum[n] = x;
         vec.push_back(n);
     }
     else
     {
         sum[n] = sum[n-] + x;
         int sz = vec.size();
         if(sum[n] > sum[vec[sz-]])
             vec.push_back(n);
     }
 }
 void D()
 {
     if(n == )return;
     int sz = vec.size();
     if(vec[sz-] == n)
         vec.pop_back();
     n--;
     int t = pre[now];
     next[t] = next[now];
     if(next[now] != -)pre[next[now]] = t;
     now = t;
 }
 void L()
 {
     if(n == )return;
     int sz = vec.size();
     if(vec[sz-] == n)
         vec.pop_back();
     now = pre[now];
     n--;
 }
 void R()
 {
     if(next[now] == -)return;
     n++;
     now = next[now];
     if(n == )
     {
         sum[n] = a[now];
         vec.push_back(n);
     }
     else
     {
         int sz = vec.size();
         sum[n] = sum[n-] + a[now];
         if(sum[n] > sum[vec[sz-]])
             vec.push_back(n);
     }
 }

 int Q(int k)
 {
     int sz = vec.size();
     if(vec[sz-] <= k)
         return sum[vec[sz-]];
     int t = upper_bound(vec.begin(),vec.end(),k) - vec.begin();
     return sum[vec[t-]];
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int M;
     int x;
     char op[];
     while(scanf("%d",&M) == )
     {
         init();
         while(M--)
         {
             scanf("%s",&op);
             if(op[] == 'I')
             {
                 scanf("%d",&x);
                 I(x);
             }
             else if(op[] == 'D')
                 D();
             else if(op[] == 'L')
                 L();
             else if(op[] == 'R')
                 R();
             else
             {
                 scanf("%d",&x);
                 printf("%d\n",Q(x));
             }
         }
     }
     return ;
 }