HDU 4699 Editor

Editor

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

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

 思路：两个栈的应用 。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <vector>
#include <set>
#include <algorithm>
#include <vector>
#include <stack>
#include <math.h>
#include <stdlib.h>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long LL ;
const int size=1000008 ;
int sum[size] ;
int Left[size] ;
int Right[size] ;
int dp[size] ;
struct Me{
    int N ;
    int left_top  ;
    int right_top ;
    Me(){} ;
    Me(int n):N(n){
        left_top=0 ;
        right_top=0 ;
        sum[0]=0 ;
        fill(dp,dp+1+N,-10000) ;
    }
    void gao(){
       char str[2] ;
       int x ;
       for(int i=1;i<=N;i++){
           scanf("%s",str) ;
           if(str[0]=='I'){
              scanf("%d",&x) ;
              Left[++left_top]=x ;
              sum[left_top]=sum[left_top-1]+x ;
              dp[left_top]=Max(dp[left_top-1],sum[left_top]) ;
           }
           else if(str[0]=='D'){
               if(left_top)
                    left_top-- ;
           }
           else if(str[0]=='L'){
               if(left_top){
                  Right[++right_top]=Left[left_top] ;
                  left_top-- ;
               }
           }
           else if(str[0]=='R'){
                if(right_top){
                    Left[++left_top]=Right[right_top] ;
                    right_top-- ;
                    sum[left_top]=sum[left_top-1]+Left[left_top] ;
                    dp[left_top]=Max(dp[left_top-1],sum[left_top]) ;
                }
           }
           else{
               scanf("%d",&x) ;
               printf("%d
",dp[Min(left_top,x)]) ;
           }
       }
    }
};
int main(){
   int n ;
   while(scanf("%d",&n)!=EOF){
      Me me(n);
      me.gao() ;
   }
   return 0 ;
}