Can you answer these queries VI
Time Limit: 2000ms
Memory Limit: 262144KB
This problem will be judged on SPOJ. Original ID: GSS664-bit integer IO format: %lld Java class name: Main
Given a sequence A of N (N <= 100000) integers, you have to apply Q (Q <= 100000) operations:
Insert, delete, replace an element, find the maximum contiguous(non empty) sum in a given interval.
Input
The first line of the input contains an integer N.
The following line contains N integers, representing the starting
sequence A1..AN, (|Ai| <= 10000).
The third line contains an integer Q. The next Q lines contains the operations in following form:
I x y: insert element y at position x (between x - 1 and x).
D x : delete the element at position x.
R x y: replace element at position x with y.
Q x y: print max{Ai + Ai+1 + .. + Aj | x <= i <= j <= y}.
All given positions are valid, and given values are between -10000 and +10000.
The sequence will never be empty.
Output
For each "Q" operation, print an integer(one per line) as described above.
Example
Input:
5
3 -4 3 -1 6
10
I 6 2
Q 3 5
R 5 -4
Q 3 5
D 2
Q 1 5
I 2 -10
Q 1 6
R 2 -1
Q 1 6
Output:
8
3
6
3
5
Source
解题：splay
1 #include <bits/stdc++.h> 2 #define KT ch[ch[root][1]][0] 3 using namespace std; 4 const int INF = numeric_limits<int>::max(); 5 const int maxn = 210010; 6 7 struct SplayTree { 8 int fa[maxn],ch[maxn][2],sz[maxn],key[maxn]; 9 int lsum[maxn],rsum[maxn],ans[maxn],sum[maxn]; 10 int tot,root,seq[maxn]; 11 inline void pushup(int x) { 12 if(!x) return; 13 sz[x] = 1 + sz[ch[x][0]] + sz[ch[x][1]]; 14 sum[x] = key[x] + sum[ch[x][0]] + sum[ch[x][1]]; 15 lsum[x] = max(lsum[ch[x][0]],key[x] + sum[ch[x][0]] + max(0,lsum[ch[x][1]])); 16 rsum[x] = max(rsum[ch[x][1]],key[x] + sum[ch[x][1]] + max(0,rsum[ch[x][0]])); 17 ans[x] = max(max(ans[ch[x][0]],ans[ch[x][1]]),key[x] + max(0,rsum[ch[x][0]]) + max(0,lsum[ch[x][1]])); 18 } 19 void newnode(int &x,int val,int f) { 20 x = ++tot; 21 lsum[x] = rsum[x] = ans[x] = sum[x] = key[x] = val; 22 fa[x] = f; 23 ch[x][0] = ch[x][1] = 0; 24 sz[x] = 1; 25 } 26 void build(int &x,int L,int R,int f) { 27 if(L > R) return; 28 int mid = (L + R)>>1; 29 newnode(x,seq[mid],f); 30 build(ch[x][0],L,mid-1,x); 31 build(ch[x][1],mid+1,R,x); 32 pushup(x); 33 } 34 void init(int n) { 35 tot = root = 0; 36 ch[0][0] = ch[0][1] = fa[0] = sum[0] = 0; 37 lsum[0] = rsum[0] = ans[0] = key[0] = -INF; 38 newnode(root,-INF,0); 39 newnode(ch[root][1],-INF,root); 40 build(KT,1,n,ch[root][1]); 41 pushup(ch[root][1]); 42 pushup(root); 43 } 44 void rotate(int x,int kd) { 45 int y = fa[x]; 46 ch[y][kd^1] = ch[x][kd]; 47 fa[ch[x][kd]] = y; 48 fa[x] = fa[y]; 49 ch[x][kd] = y; 50 fa[y] = x; 51 if(fa[x]) ch[fa[x]][y == ch[fa[x]][1]] = x; 52 pushup(y); 53 } 54 void splay(int x,int goal = 0) { 55 while(fa[x] != goal) { 56 if(fa[fa[x]] == goal) rotate(x,x == ch[fa[x]][0]); 57 else { 58 int y = fa[x],z = fa[y],s = (y == ch[z][0]); 59 if(x == ch[y][s]) { 60 rotate(x,s^1); 61 rotate(x,s); 62 } else { 63 rotate(y,s); 64 rotate(x,s); 65 } 66 } 67 } 68 pushup(x); 69 if(!goal) root = x; 70 } 71 int select(int k,int goal) { 72 int x = root; 73 while(sz[ch[x][0]] + 1 != k) { 74 if(k < sz[ch[x][0]] + 1) x = ch[x][0]; 75 else { 76 k -= sz[ch[x][0]] + 1; 77 x = ch[x][1]; 78 } 79 } 80 splay(x,goal); 81 return x; 82 } 83 void insert(int a,int b) { 84 select(a - 1 + 1,0); 85 select(a + 1,root); 86 newnode(KT,b,ch[root][1]); 87 pushup(ch[root][1]); 88 pushup(root); 89 } 90 void remove(int a) { 91 select(a - 1 + 1,0); 92 select(a + 1 + 1,root); 93 KT = 0; 94 pushup(ch[root][1]); 95 pushup(root); 96 } 97 void replace(int a,int b){ 98 int x = root; 99 ++a; 100 while(sz[ch[x][0]] + 1 != a){ 101 if(a < sz[ch[x][0]] + 1) x = ch[x][0]; 102 else{ 103 a -= sz[ch[x][0]] + 1; 104 x = ch[x][1]; 105 } 106 } 107 key[x] = b; 108 splay(x,0); 109 } 110 int query(int a,int b){ 111 select(a-1+1,0); 112 select(b+1+1,root); 113 return ans[KT]; 114 } 115 } spt; 116 int main() { 117 int n,m,x,y; 118 char op[10]; 119 while(~scanf("%d",&n)) { 120 for(int i = 1; i <= n; ++i) 121 scanf("%d",&spt.seq[i]); 122 spt.init(n); 123 scanf("%d",&m); 124 while(m--) { 125 scanf("%s",op); 126 if(op[0] == 'I') { 127 scanf("%d%d",&x,&y); 128 spt.insert(x,y); 129 } else if(op[0] == 'D') { 130 scanf("%d",&x); 131 spt.remove(x); 132 }else if(op[0] == 'R'){ 133 scanf("%d%d",&x,&y); 134 spt.replace(x,y); 135 }else if(op[0] == 'Q'){ 136 scanf("%d%d",&x,&y); 137 printf("%d ",spt.query(x,y)); 138 } 139 } 140 } 141 return 0; 142 }