题目链接: http://poj.org/problem?id=3468
题意:区间更新,求区间和。
本来是线段树区间更新的入门题,一直听说Splay这个神奇的数据结构没去学,于是就学了一下,第一道。。
推荐学习:
http://notonlysuccess.me/?p=709
http://blog.youkuaiyun.com/acm_cxlove/article/details/7815019
先主要搞懂两种旋转方式,实现就是Rotate和Splay两个操作,这几基本一样没什么变化。
主要变化就是区间操作的时候,要加入lazy标记,然后添加了pushUp和pushDown两个操作。
#include <iostream>
#include <cstdio>
using namespace std;
#define Key_value (ch[ch[root][1]][0])
typedef long long ll;
const int maxn = 1e5+10;
int n, q, a[maxn];
// splay
int tot1, root;
int sz[maxn], pre[maxn], ch[maxn][2];
ll add[maxn], sum[maxn], val[maxn];
//debug部分copy from hh
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 , sum = %2d \n",x,ch[x][0],ch[x][1],pre[x],sz[x],val[x],sum[x]);
Treaval(ch[x][1]);
}
}
void debug() {printf("%d\n",root);Treaval(root);}
void newNode(int &r, int father, int k)
{
r = ++tot1;
pre[r] = father;
sz[r] = 1;
val[r] = k;
sum[r] = k;
add[r] = 0;
ch[r][0]= ch[r][1] = 0;
}
void pushUp(int r)
{
sz[r] = sz[ch[r][0]] + sz[ch[r][1]] + 1;
sum[r] = sum[ch[r][0]] + sum[ch[r][1]] + val[r] + add[r];
}
void pushDown(int r)
{
if(add[r])
{
val[r] += add[r];
add[ch[r][0]] += add[r];
add[ch[r][1]] += add[r];
sum[ch[r][0]] += 1LL*add[r]*sz[ch[r][0]];
sum[ch[r][1]] += 1LL*add[r]*sz[ch[r][1]];
add[r] = 0;
}
}
void build(int &x, int l, int r, int father)
{
if(l > r) return;
int mid = (l + r) >> 1;
newNode(x, father, a[mid]);
if(l < mid) build(ch[x][0], l, mid-1, x);
if(r > mid) build(ch[x][1], mid+1, r, x);
pushUp(x);
}
// kind=0左旋, kind=1右旋
void Rotate(int x, int kind)
{
int y = pre[x];
pushDown(x);
pushDown(y);
//
ch[y][!kind] = ch[x][kind];
pre[ch[x][kind]] = y;
if(pre[y]) ch[pre[y]][y == ch[pre[y]][1]] = x;
pre[x] = pre[y];
pre[y] = x;
ch[x][kind] = y;
pushUp(y);
}
void Splay(int r, int goal)
{
pushDown(r);
while(pre[r] != goal)
{
if(pre[pre[r]] == goal) Rotate(r, r == ch[pre[r]][0]);
else
{
int y = pre[r];
int kind = ch[pre[y]][0] == y;
// 方向不同
if(ch[y][kind] == r)
{
Rotate(r, !kind);
Rotate(r, kind);
}
// 方向相同,相同方向转两次
else
{
Rotate(y, kind);
Rotate(r, kind);
}
}
}
pushUp(r);
if(goal == 0) root = r;
}
// 把第k位的数转到goal下面
void RotateTo(int k, int goal)
{
int r = root;
pushDown(r);
while(sz[ch[r][0]] != k)
{
if(k < sz[ch[r][0]]) r = ch[r][0];
else
{
k -= ( sz[ch[r][0]] + 1);
r = ch[r][1];
}
pushDown(r);
}
Splay(r, goal);
}
ll query(int l, int r)
{
RotateTo(l-1, 0);
//debug();
RotateTo(r+1, root);
//debug();
return sum[Key_value];
}
void update(int l, int r, int c)
{
RotateTo(l-1, 0);
RotateTo(r+1, root);
add[Key_value] += c;
sum[Key_value] += 1LL*sz[Key_value] * c;
}
void init()
{
for(int i=0;i<n;i++) scanf("%d", &a[i]);
sz[0] = ch[0][0] = ch[0][1] = pre[0] = 0;
add[0] = sum[0] = 0;
root = tot1 = 0;
newNode(root, 0, -1);
newNode(ch[root][1], root, -1);
sz[root] = 2;
build(Key_value, 0 , n-1, ch[root][1]);
pushUp(ch[root][1]);
pushUp(root);
}
int main()
{
while(~scanf("%d%d", &n, &q))
{
init();
//debug();
while(q--)
{
char buf[10];
int x, y, c;
scanf("%s%d%d", buf, &x, &y);
if(buf[0] == 'Q') printf("%lld\n", query(x, y));
else
{
scanf("%d", &c);
update(x, y, c);
}
}
}
return 0;
}