Description
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4
Sample Output
4 55 9 15
Hint
The sums may exceed the range of 32-bit integers.
Source
POJ Monthly--2007.11.25, Yang Yi
【分析】
这个其实是个水题。我只是拿来fhq treap的...
Orzzzzzz范大爷
这种数据结构真是...能保证treap的性质同时也能保证合并后的树还能原封不动的切出来...
说它是treap，我觉得就一个fix像treap而已。
这个数据结构的精髓在合并和分裂(和打标记？)。有注释.
1 #include <iostream>
2 #include <cstdio>
3 #include <algorithm>
4 #include <cstring>
5 #include <vector>
6 #include <utility>
7 #include <iomanip>
8 #include <string>
9 #include <cmath>
10 #include <queue>
11 #include <assert.h>
12 #include <map>
13 #include <ctime>
14 #include <cstdlib>
15 #define LOCAL
16 const int MAXN = 100000 + 10;
17 const int INF = 0x3f3f3f3f;
18 const int SIZE = 450;
19 const int maxnode = 1101000;
20 using namespace std;
21 typedef long long ll;
22 ll n, m;
23 struct fhqTreap{
24 struct Node{
25 Node *l, *r;
26 ll delta, sum;
27 ll size, fix, val;
28 }*root, *null, mem[1101000];
29 ll tot;
30
31 ll BIG_RAND(){return (ll)rand()*RAND_MAX + (ll)rand();}
32 void init(){
33 tot = 0;
34 NEW(null, 0);
35 null->size = 0;
36 root = null;
37 insert(0, n);//插入
38 }
39 void update(Node *t){
40 if (t == null) return;
41 t->size = t->l->size + t->r->size + 1;
42 t->sum = t->val;
43 if (t->l != null) t->sum += t->l->sum;
44 if (t->r != null) t->sum += t->r->sum;
45 }
46 //标记下传=-=
47 void push_down(Node *t){
48 if (t->delta){
49 t->val += t->delta;
50
51 if (t->l != null) {t->l->delta += t->delta, t->l->sum += t->l->size * t->delta;}
52 if (t->r != null) {t->r->delta += t->delta, t->r->sum += t->r->size * t->delta;}
53
54 t->delta = 0;
55 }
56 }
57
58 void NEW(Node *&t, ll val){
59 t = &mem[tot++];
60 t->size = 1;
61 t->fix = BIG_RAND();
62 t->sum = t->val = val;
63 t->delta = 0;
64 t->l = t->r = null;
65 }
66 //将t分裂为大小p和t->size - p的两个子树
67 void split(Node *t, Node *&a, Node *&b, int p){
68 if (t->size <= p) a = t, b = null;
69 else if (p == 0) a = null, b = t;
70 else {
71 push_down(t);
72 if (t->l->size >= p){
73 b = t;
74 split(t->l, a , b->l, p);
75 update(b);
76 }else{
77 a = t;
78 split(t->r, a->r, b, p - t->l->size - 1);
79 update(a);
80 }
81 }
82 }
83 //将a,b树合并为t
84 void merge(Node *&t, Node *a, Node *b){
85 if (a == null) t = b;
86 else if (b == null) t = a;
87 else {
88 if (a->fix > b->fix){//按fix值排序
89 push_down(a);
90 t = a;
91 merge(t->r, a->r, b);
92 }else{
93 push_down(b);
94 t = b;
95 merge(t->l, a, b->l);
96 }
97 update(t);
98 }
99 }
100 void add(ll l, ll r, ll val){
101 Node *a, *b, *c;
102 split(root, a, b, l - 1);
103 split(b, b, c, r - l + 1);
104 b->delta += val;
105 b->sum += val * b->size;
106 merge(a, a, b);
107 merge(root, a, c);
108 }
109 ll query(ll l, ll r){
110 Node *a, *b, *c;
111 split(root, a, b, l - 1);
112 split(b, b, c, r - l + 1);
113 ll ans = b->sum;
114 merge(b, b, c);
115 merge(root, a, b);
116 return ans;
117 }
118 //把b挤出去
119 void Delete(ll p){
120 Node *a, *b, *c;
121 split(root, a, b, p - 1);
122 split(b, b, c, 1);
123 merge(root, a, c);
124 }
125 Node *Insert(ll x){//不可思议的插入方式
126 if (x == 0) return null;
127 if (x == 1){
128 ll tmp;
129 scanf("%lld", &tmp);
130 Node *a;
131 NEW(a, tmp);
132 return a;
133 }
134 Node *a, *b;
135 int mid = x / 2;
136 a = Insert(mid);
137 b = Insert(x - mid);
138 merge(a, a, b);
139 return a;
140 }
141
142 //在pos处插入x个数字
143 void insert(ll pos, ll x){
144 Node *a, *b, *c, *t;
145 //跟块状链表的有点像，分裂再合并
146 split(root, a, b, pos);
147 c = Insert(x);//插入一个值为x的树
148 merge(a, a, c);
149 merge(root, a, b);
150 }
151 }A;
152
153 void work(){
154 char str[3];
155 while (m--){
156 scanf("%s", str);
157 if (str[0] == ‘Q‘){
158 ll l, r;
159 scanf("%lld%lld", &l, &r);
160 printf("%lld\n", A.query(l, r));
161 }else{
162 ll l, r, k;
163 scanf("%lld%lld%lld", &l, &r, &k);
164 A.add(l, r, k);
165 }
166 }
167 }
168
169 int main(){
170
171 while( scanf("%lld%lld", &n, &m) != EOF){
172 A.init();
173 work();
174 }
175 return 0;
176 }