hdu 4699 Editor (巧用两个栈)

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题目：http://acm.hdu.edu.cn/showproblem.php?pid=4699

题意：

Editor

Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2306 Accepted Submission(s): 704

Problem Description

Sample Input

8I 2I -1I 1Q 3LDRQ 2

Sample Output

23Hint
The following diagram shows the status of sequence after each instruction:

Source

2013 Multi-University Training Contest 10

分析：用两个栈存数，光标始终在第一个栈的栈顶，左移s1就向s2里面弹一个数，右移s2就向s1里面弹一个值，插入和删除直接在s1的栈顶操作。找最大前缀和，可以用两个数组维护，一个记录前缀和，一个记录最大前缀和，当增加一个数，最大前缀和maxpre[k]只可能是maxpre[k-1]和pre[k]里面的最大值。

代码：

#include <iostream>#include <cstring>#include <cstdio>#include <stack>using namespace std;const int maxn = 1e6+7;const int INF = 2E9;int pre[maxn],maxpre[maxn],cur,x,q;stack <int > s1,s2;void Init(){maxpre[0]=pre[0]=-INF;cur=1;while(!s1.empty())s1.pop();while(!s2.empty())s2.pop();}void Insert(){scanf("%d",&x);s1.push(x);pre[cur]=pre[cur-1]+x;if(cur==1)pre[cur]=x;maxpre[cur]=max(maxpre[cur-1],pre[cur]);cur++;}void Delete(){if(s1.empty())return ;s1.pop();cur--;}void Lmove(){if(s1.empty())return ;int temp=s1.top();s1.pop();s2.push(temp);cur--;}void Rmove(){if(s2.empty())return ;int temp=s2.top();s2.pop();s1.push(temp);pre[cur]=pre[cur-1]+temp;if(cur==1)pre[cur]=temp;maxpre[cur]=max(maxpre[cur-1],pre[cur]);cur++;}void Query(){scanf("%d",&x);printf("%d\n",maxpre[x]);}int main(){char str[3];while(scanf("%d",&q)!=EOF){Init();while(q--){scanf("%s",str);switch(str[0]){case 'I':Insert();break;case 'D':Delete();break;case 'L':Lmove();break;case 'R':Rmove();break;case 'Q':Query();break;default:break;}}}return 0;}

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