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！
1 #include<stack> 2 #include<cstdio> 3 #include<cstring> 4 #define MAX 1000005 5 using namespace std; 6 stack<int>l, r; 7 long long int dp[MAX], sum[MAX], n, temp; 8 char str[5]; 9 int main(){ 10 freopen("in.c", "r", stdin); 11 while(~scanf("%lld", &n)){ 12 while(!l.empty()) l.pop(); 13 while(!r.empty()) r.pop(); 14 memset(sum, 0, sizeof(sum)); 15 memset(dp, 0, sizeof(dp)); 16 dp[0] = -100000000; 17 for(int i = 0;i < n;i ++){ 18 scanf("%s", str); 19 if(str[0] == 'I'){ 20 scanf("%lld", &temp); 21 l.push(temp); 22 sum[l.size()] = sum[l.size()-1] + temp; 23 dp[l.size()] = max(sum[l.size()], dp[l.size()-1]); 24 }else if(str[0] == 'L'){ 25 if(!l.empty()){ 26 int del = l.top(); 27 l.pop(); 28 r.push(del); 29 } 30 }else if(str[0] == 'Q'){ 31 scanf("%lld", &temp); 32 temp = min(temp, (long long int)l.size()); 33 l.size() == 0 ? printf("0\n") : printf("%lld\n", dp[temp]); 34 }else if(str[0] == 'D') l.pop(); 35 else{ 36 if(!r.empty()){ 37 int del = r.top(); 38 r.pop(); 39 l.push(del); 40 sum[l.size()] = sum[l.size()-1] + del; 41 dp[l.size()] = max(sum[l.size()], dp[l.size()-1]); 42 } 43 } 44 } 45 } 46 }
