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
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101004.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 ",Q(x)); 146 } 147 } 148 } 149 return 0; 150 }