HDOJ -- 4699

Editor

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1236    Accepted Submission(s): 391

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:

 

 

一直wa，谁能告诉我哪错了！

 

 #include<stack>
 #include<cstdio>
 #include<cstring>
 #define MAX 1000005
 using namespace std;
 stack<int>l, r;
 long long int dp[MAX], sum[MAX], n, temp;
 char str[];
 int main(){
     freopen("in.c", "r", stdin);
     while(~scanf("%lld", &n)){
         while(!l.empty()) l.pop();
         while(!r.empty()) r.pop();
         memset(sum, , sizeof(sum));
         memset(dp, , sizeof(dp));
         dp[] = -;
         for(int i = ;i < n;i ++){
             scanf("%s", str);
             if(str[] == 'I'){
                 scanf("%lld", &temp);
                 l.push(temp);
                 sum[l.size()] = sum[l.size()-] + temp;
                 dp[l.size()] = max(sum[l.size()], dp[l.size()-]);
             }else if(str[] == 'L'){
                 if(!l.empty()){
                     int del = l.top();
                     l.pop();
                     r.push(del);
                 }
             }else if(str[] == 'Q'){
                 scanf("%lld", &temp);
                 temp = min(temp, (long long int)l.size());
                 l.size() ==  ? printf("0\n") : printf("%lld\n", dp[temp]);
             }else if(str[] == 'D') l.pop();
             else{
                 if(!r.empty()){
                     int del = r.top();
                     r.pop();
                     l.push(del);
                     sum[l.size()] = sum[l.size()-] + del;
                     dp[l.size()] = max(sum[l.size()], dp[l.size()-]);
                 }
             }
         }
     }
 }