题目链接：点我~~
题意：给n个数，有两种操作，一种是查询区间和，另一种是在区间上每一个数加上v。
思路：第一次摸splay tree，这题算是个模板题，适合思考人生。。。
//Splay(x,0); 将x变为跟节点
//Splay(x,root); 将x变为root下的节点 维护区间时通常旋转为root的右子树
//所需要维护的区间[l，r]，通过旋转后即为 root右孩子的左子树
模板参考链接：ˋ( ° ▽、° ) ￣へ￣
//#include <bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<iostream>
#include<queue>
#include<cmath>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PI;
typedef pair< PI, int> PII;
const double eps=1e-5;
const double pi=acos(-1.0);
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
#define Key_value ch[ch[root][1]][0]
const int MAXN = 300000;
const int MAXM = 20100;
int pre[MAXN],ch[MAXN][2],key[MAXN],size[MAXN];
int root,tot1;
LL sum[MAXN];
int s[MAXN],tot2;//内存池和容量
int a[MAXN];
int n,q;
int add[MAXN];
void NewNode(int &r,int father,int k)
{
if(tot2) r = s[tot2--];//取的时候是tot2--,存的时候就是++tot2
else r = ++tot1;
pre[r] = father;
ch[r][0] = ch[r][1] = 0;
key[r] = k;
sum[r] = k;
size[r] = 1;
}
void Update_add(int r,int v)
{
if(!r)return;
add[r] += v;
key[r] += v;
sum[r] += (LL)v*size[r];
}
void push_up(int r)
{
int lson = ch[r][0], rson = ch[r][1];
size[r] = size[lson] + size[rson] + 1;
sum[r] = sum[lson] + sum[rson] + key[r];
}
void push_down(int r)
{
if(add[r])
{
int lson = ch[r][0], rson = ch[r][1];
Update_add(lson,add[r]);
Update_add(rson,add[r]);
add[r]=0;
}
}
void Build(int &x,int l,int r,int father)
{
if(l > r)return;
int mid = (l+r)/2;
NewNode(x,father,a[mid]);
Build(ch[x][0],l,mid-1,x);
Build(ch[x][1],mid+1,r,x);
push_up(x);
}
void Init()
{
root = tot1 = tot2 = 0;
ch[root][0] = ch[root][1] = size[root] = pre[root] = 0;
sum[root] = key[root] = 0;
NewNode(root,0,-1);
NewNode(ch[root][1],root,-1);
for(int i = 1; i <= n; i++)
scanf("%d",&a[i]);
Build(Key_value,1,n,ch[root][1]);
push_up(ch[root][1]);
push_up(root);
}
//旋转,0为左旋，1为右旋
void Rotate(int x,int kind)
{
int y = pre[x];
push_down(y);
push_down(x);
ch[y][!kind] = ch[x][kind];
pre[ch[x][kind]] = y;
if(pre[y])
ch[pre[y]][ch[pre[y]][1]==y] = x;
pre[x] = pre[y];
ch[x][kind] = y;
pre[y] = x;
push_up(y);
}
//Splay调整，将r结点调整到goal下面
void Splay(int r,int goal)
{
push_down(r);
while(pre[r] != goal)
{
if(pre[pre[r]] == goal)
{
push_down(pre[r]);
push_down(r);
Rotate(r,ch[pre[r]][0] == r);
}
else
{
push_down(pre[pre[r]]);
push_down(pre[r]);
push_down(r);
int y = pre[r];
int kind = ch[pre[y]][0]==y;
if(ch[y][kind] == r)
{
Rotate(r,!kind);
Rotate(r,kind);
}
else
{
Rotate(y,kind);
Rotate(r,kind);
}
}
}
push_up(r);
if(goal == 0) root = r;
}
int Get_kth(int r,int k) //得到第k个结点
{
push_down(r);
int t = size[ch[r][0]] + 1;
if(t == k)return r;
if(t > k)return Get_kth(ch[r][0],k);
else return Get_kth(ch[r][1],k-t);
}
//因为加了个空结点，所以将第l个点旋转到根结点，第r+2个点旋转到根结点的右孩子
//l-1 ~~ r+1
LL Get_Sum(int l,int r)
{
Splay(Get_kth(root,l),0);
Splay(Get_kth(root,r+2),root);
return sum[Key_value];
}
void modify(int l,int r,int v)
{
Splay(Get_kth(root,l),0);
Splay(Get_kth(root,r+2),root);
Update_add(Key_value,v);
push_up(ch[root][1]);
push_up(root);
}
void Treavel(int x)
{
if(x)
{
Treavel(ch[x][0]);
printf("结点：%2d: 左儿子 %2d 右儿子 %2d 父结点 %2d size\
= %2d\n",x,ch[x][0],ch[x][1],pre[x],size[x]);
Treavel(ch[x][1]);
}
}
void debug()
{
printf("root:%d\n",root);
Treavel(root);
}
int main()
{
scanf("%d%d",&n,&q);
Init();
char op[20];
int x,y,z;
while(q--)
{
scanf("%s",op);
int l,r,v;
if(op[0]=='Q')
{
scanf("%d%d",&l,&r);
cout<<Get_Sum(l,r)<<endl;;
}
else
{
scanf("%d%d%d",&l,&r,&v);
modify(l,r,v);
}
}
return 0;
}
==第二版==
//#include <bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<iostream>
#include<queue>
#include<cmath>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PI;
typedef pair< PI, int> PII;
const double eps=1e-5;
const double pi=acos(-1.0);
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
#define Key_value ch[ch[root][1]][0]
#define keyTree (ch[ch[root][1]][0])
const int MAXN = 300000;
const int MAXM = 20100;
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];
LL sum[MAXN];
inline void Rotate(int x,int f) //旋转,0为左旋，1为右旋
{
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) //将x旋转到goal的下面
{
push_down(x); //防止pre[x]就是目标点，下面的循环就进不去了，x的信息就传不下去了
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);
}
inline int Get_kth(int r,int k) //得到第k个结点
{
push_down(r);
int t = sz[ch[r][0]] + 1;
if(t == k)return r;
if(t > k)return Get_kth(ch[r][0],k);
else return Get_kth(ch[r][1],k-t);
}
//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 类似线段树
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] ] += (LL)sz[ ch[x][0] ] * add[x];
sum[ ch[x][1] ] += (LL)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);
}
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; //sz当前节点儿子的数量
sum[keyTree] += (LL)c * sz[ keyTree ];
}
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[5];
scanf("%s",op);
if(op[0] == 'Q')
{
spt.query();
}
else
{
spt.update();
//spt.debug();
}
}
return 0;
}
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