poj 3468 A Simple Problem with Integers(splay tree模板题)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 36775		Accepted: 10613
Case Time Limit: 2000MS

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.

这俩篇论文很好：http://www.docin.com/p-62465596.html

                               http://www.docin.com/p-63165342.html

模板留着用：出处http://www.notonlysuccess.com/index.php/splay-tree/

建树时是按照数列序号从小到大排好的，每个节点左子树的序号小于右子树的序号及这个节点本身，以后每次旋转也是在这个基础上..

注意：因为询问时要伸展l-1和r+1两个节点，所以我们就有必要在原数列的头尾都再加上一个任意数，这样就能保证一定能找到l-1和r+1。因为数列中第一个数前面没有数字了，并且最后一个数后面也没有数字了，这样提取区间时就会出一些问题。为了不进行过多的特殊判断， 我们在原数列最前面和最后面分别加上一个数， 在伸展树中就体现为结点，这样提取区间的时候原来的第 k 个数就是现在的第k+1 个数。并且我们还要注意，这两个结点维护的信息不能影响到正确的结果。

在这题里另加的这俩个节点的val值可以是任意数都不影响正确的求和结果，因为当提取区间[a,b]时，将a前面一个数对应的结点转到树根后， a及a后面的数就在根的右子树上，然后又将b 后面一个结点对应的结点转到树根的右边，那么[a,b]这个区间就是下图中*所示的子树。在*子树中没有这俩个节点存在，所以不会影响到正确结果。。。。。

 如果有什么不正确的地方欢迎各位大牛指正。。。。。欢迎讨论哦！！

#include <cstdio>
#define keyTree (ch[ ch[root][1] ][0])
const int maxn = 222222;
struct SplayTree{
int sz[maxn];
int ch[maxn][2];
int pre[maxn];
int root , top1;
/*这是题目特定变量*/
int num[maxn];
int val[maxn];
int add[maxn];
long long sum[maxn];
inline void Rotate(int x,int f) {
int y = pre[x];
push_down(y);
push_down(x);
ch[y][!f] = ch[x][f];
pre[ ch[x][f] ] = y;
pre[x] = pre[y];
if(pre[x]) ch[ pre[y] ][ ch[pre[y]][1] == y ] = x;
ch[x][f] = y;
pre[y] = x;
push_up(y);
}
inline void Splay(int x,int goal) {
push_down(x);
while(pre[x] != goal) {
if(pre[pre[x]] == goal) {
Rotate(x , ch[pre[x]][0] == x);
} else {
int y = pre[x] , z = pre[y];
int f = (ch[z][0] == y);
if(ch[y][f] == x) {
Rotate(x , !f) , Rotate(x , f);
} else {
Rotate(y , f) , Rotate(x , f);
}
}
}
push_up(x);
if(goal == 0) root = x;
}
inline void RotateTo(int k,int goal) {//把第k位的数转到goal下边
int x = root;
push_down(x);
while(sz[ ch[x][0] ] != k) {
if(k < sz[ ch[x][0] ]) {
x = ch[x][0];
} else {
k -= (sz[ ch[x][0] ] + 1);//多减去1，所以k+1位
x = ch[x][1];
}
push_down(x);
}
Splay(x,goal);
}
//以上一般不修改//////////////////////////////////////////////////////////////////////////////
void debug() {printf("%d\n",root);Treaval(root);}
void Treaval(int x) {
if(x) {
Treaval(ch[x][0]);
printf("结点%2d:左儿子 %2d 右儿子 %2d 父结点 %2d size = %2d ,val = %2d\n",x,ch[x][0],ch[x][1],pre[x],sz[x],val[x]);
Treaval(ch[x][1]);
}
}
//以上Debug

//以下是题目的特定函数:
inline void NewNode(int &x,int c) {
x = ++top1;
ch[x][0] = ch[x][1] = pre[x] = 0;
sz[x] = 1;

val[x] = sum[x] = c;/*这是题目特定函数*/
add[x] = 0;
}

//把延迟标记推到孩子
inline void push_down(int x) {/*这是题目特定函数*/
if(add[x]) {
val[x] += add[x];
add[ ch[x][0] ] += add[x];
add[ ch[x][1] ] += add[x];
sum[ ch[x][0] ] += (long long)sz[ ch[x][0] ] * add[x];
sum[ ch[x][1] ] += (long long)sz[ ch[x][1] ] * add[x];
add[x] = 0;
}
}
//把孩子状态更新上来
inline void push_up(int x) {
sz[x] = 1 + sz[ ch[x][0] ] + sz[ ch[x][1] ];
/*这是题目特定函数*/
sum[x] = add[x] + val[x] + sum[ ch[x][0] ] + sum[ ch[x][1] ];
}

/*初始化*/
inline void makeTree(int &x,int l,int r,int f) {
if(l > r) return ;
int m = (l + r)>>1;
NewNode(x , num[m]);		/*num[m]权值改成题目所需的*/
makeTree(ch[x][0] , l , m - 1 , x);
makeTree(ch[x][1] , m + 1 , r , x);
pre[x] = f;
push_up(x);
}
inline void init(int n) {/*这是题目特定函数*/
ch[0][0] = ch[0][1] = pre[0] = sz[0] = 0;
add[0] = sum[0] = 0;

root = top1 = 0;
//为了方便处理边界,加两个边界顶点
NewNode(root , 0);
NewNode(ch[root][1] , 0);
pre[top1] = root;
sz[root] = 2;
for (int i = 0 ; i < n ; i ++) scanf("%d",&num[i]);
makeTree(keyTree , 0 , n-1 , ch[root][1]);
push_up(ch[root][1]);
push_up(root);
}
/*更新*/
inline void update( ) {/*这是题目特定函数*/
int l , r , c;
scanf("%d%d%d",&l,&r,&c);
RotateTo(l-1,0);
RotateTo(r+1,root);
add[ keyTree ] += c;
sum[ keyTree ] += (long long)c * sz[ keyTree ];
}
/*询问*/
inline void query() {/*这是题目特定函数*/
int l , r;
scanf("%d%d",&l,&r);
RotateTo(l-1 , 0);
RotateTo(r+1 , root);
printf("%lld\n",sum[keyTree]);
}
}spt;
int main() {
int n , m;
scanf("%d%d",&n,&m);
spt.init(n);
//spt.debug();
while(m --) {
char op[2];
scanf("%s",op);
if(op[0] == 'Q') {
spt.query();
} else {
spt.update();
//spt.debug();
}
}
return 0;
}