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。

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013/8/22 13:38:35
  4 File Name     :F:\2013ACM练习\2013多校10\1004.cpp
  5 ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 const int MAXN = 1000010;
 21 
 22 int a[MAXN];
 23 int next[MAXN],pre[MAXN],tot;
 24 int NewNode()
 25 {
 26     next[tot] = -1;
 27     pre[tot] = -1;
 28     tot++;
 29     return tot-1;
 30 }
 31 int root;
 32 int sum[MAXN];
 33 vector<int>vec;
 34 int now;
 35 int n;
 36 void init()
 37 {
 38     tot = 0;
 39     root = NewNode();
 40     vec.clear();
 41     n = now = 0;
 42 }
 43 
 44 void I(int x)
 45 {
 46     n++;
 47     int t = NewNode();
 48     a[t] = x;
 49     pre[t] = now;
 50     next[t] = next[now];
 51     if(next[now] != -1)pre[next[now]] = t;
 52     next[now] = t;
 53     now = t;
 54     if(n == 1)
 55     {
 56         sum[n] = x;
 57         vec.push_back(n);
 58     }
 59     else
 60     {
 61         sum[n] = sum[n-1] + x;
 62         int sz = vec.size();
 63         if(sum[n] > sum[vec[sz-1]])
 64             vec.push_back(n);
 65     }
 66 }
 67 void D()
 68 {
 69     if(n == 0)return;
 70     int sz = vec.size();
 71     if(vec[sz-1] == n)
 72         vec.pop_back();
 73     n--;
 74     int t = pre[now];
 75     next[t] = next[now];
 76     if(next[now] != -1)pre[next[now]] = t;
 77     now = t;
 78 }
 79 void L()
 80 {
 81     if(n == 0)return;
 82     int sz = vec.size();
 83     if(vec[sz-1] == n)
 84         vec.pop_back();
 85     now = pre[now];
 86     n--;
 87 }
 88 void R()
 89 {
 90     if(next[now] == -1)return;
 91     n++;
 92     now = next[now];
 93     if(n == 1)
 94     {
 95         sum[n] = a[now];
 96         vec.push_back(n);
 97     }
 98     else
 99     {
100         int sz = vec.size();
101         sum[n] = sum[n-1] + a[now];
102         if(sum[n] > sum[vec[sz-1]])
103             vec.push_back(n);
104     }
105 }
106 
107 
108 int Q(int k)
109 {
110     int sz = vec.size();
111     if(vec[sz-1] <= k)
112         return sum[vec[sz-1]];
113     int t = upper_bound(vec.begin(),vec.end(),k) - vec.begin();
114     return sum[vec[t-1]];
115 }
116 
117 
118 int main()
119 {
120     //freopen("in.txt","r",stdin);
121     //freopen("out.txt","w",stdout);
122     int M;
123     int x;
124     char op[10];
125     while(scanf("%d",&M) == 1)
126     {
127         init();
128         while(M--)
129         {
130             scanf("%s",&op);
131             if(op[0] == 'I')
132             {
133                 scanf("%d",&x);
134                 I(x);
135             }
136             else if(op[0] == 'D')
137                 D();
138             else if(op[0] == 'L')
139                 L();
140             else if(op[0] == 'R')
141                 R();
142             else
143             {
144                 scanf("%d",&x);
145                 printf("%d\n",Q(x));
146             }
147         }
148     }
149     return 0;
150 }