poj 3468【线段树】

题意：给定Q（1<=Q<=100000）个数A1，A2…AQ，以及可能多次进行的两个操作

1）对某个区间Ai……Aj的每个数都加n（n可变）

2）对某个区间Ai……Aj的数求和

分析：

树结点只存和，会导致每次加数时都要更新到叶子节点，速度太慢(O(nlog(n)))，这是必须避免的

1.在增加时，如果要加的区间正好覆盖一个节点，则增加其节点的Inc值，不再往下走，否则要更新nSum（加上本次增量）

再将增量往下传。这样更新的复杂度就是O(log(n))

2.在查询时，如果待查区间不是正好覆盖一个节点，就将节点的Inc往下带，然后将Inc代表的所有增量累加到nSum上后将

Inc清0，接下来再往下查询。Inc往下带的过程也是区间分解过程，复杂度也是O(log(n))

#include<algorithm>
#include<cstdio>
#include<vector>
#include<string>
#include<string.h>
#include<iostream>
using namespace std;
typedef long long LL;
const int INF = 0x7FFFFFFF;
const int maxn = 1e3 + 10;

struct CNode
{
	int L, R;
	CNode *pLeft, *pRight;
	long long nSum;//原来的和
	long long Inc;//增量c的累加
};

CNode Tree[200010];//2倍叶子节点数目就够
int nCount = 0;
int Mid(CNode*pRoot)
{
	return (pRoot->L + pRoot->R) / 2;
}

void BuildTree(CNode *pRoot, int L, int R)
{
	pRoot->L = L;
	pRoot->R = R;
	pRoot->nSum = 0;
	pRoot->Inc = 0;
	if (L == R)
		return;
	nCount++;
	pRoot->pLeft = Tree + nCount;
	nCount++;
	pRoot->pRight = Tree + nCount;
	BuildTree(pRoot->pLeft, L, (L + R) / 2);
	BuildTree(pRoot->pRight, (L + R) / 2 + 1, R);
}

void Insert(CNode *pRoot, int i, int v)
{
	if (pRoot->L == i&&pRoot->R == i)
	{
		pRoot->nSum = v;
		return;
	}
	pRoot->nSum += v;//累加和
	if (i <= Mid(pRoot))
		Insert(pRoot->pLeft, i, v);
	else
		Insert(pRoot->pRight, i, v);

}

void Add(CNode * pRoot, int a, int b, long long c)
{
	if (pRoot->L == a&&pRoot->R == b)
	{
		pRoot->Inc += c;
		return;
	}
	pRoot->nSum += c*(b - a + 1);
	if (b <= (pRoot->L + pRoot->R) / 2)
		Add(pRoot->pLeft, a, b, c);
	else if (a >= (pRoot->L + pRoot->R) / 2 + 1)
		Add(pRoot->pRight, a, b, c);
	else
	{
		Add(pRoot->pLeft, a, (pRoot->L + pRoot->R) / 2, c);
		Add(pRoot->pRight, (pRoot->L + pRoot->R) / 2 + 1, b, c);
	}

}

long long QuerynSum(CNode * pRoot, int a, int b)
{
	if (pRoot->L == a&&pRoot->R == b)
		return pRoot->nSum + (pRoot->R - pRoot->L + 1)*pRoot->Inc;
	pRoot->nSum += (pRoot->R - pRoot->L + 1)*pRoot->Inc;
	Add(pRoot->pLeft, pRoot->L, Mid(pRoot), pRoot->Inc);
	Add(pRoot->pRight, Mid(pRoot) + 1, pRoot->R, pRoot->Inc);
	pRoot->Inc = 0;

	if (b <= Mid(pRoot))
		return QuerynSum(pRoot->pLeft, a, b);
	else if (a >= Mid(pRoot) + 1)
		return QuerynSum(pRoot->pRight, a, b);
	else
	{
		return QuerynSum(pRoot->pLeft, a, Mid(pRoot)) +
               QuerynSum(pRoot->pRight, Mid(pRoot) + 1, b);
	}
}

int main()
{
	int n, q, a, b, c;
	char cmd[10];
	scanf("%d%d", &n, &q);
	int i, j, k;
	nCount = 0;
	BuildTree(Tree, 1, n);
	for (i = 1; i <= n;i++)
	{
		scanf("%d", &a);
		Insert(Tree, i, a);
	}
	for (i = 0; i < q; i++)
	{
		scanf("%s", cmd);
		if (cmd[0]=='C')
		{
			scanf("%d%d%d", &a, &b, &c);
			Add(Tree, a, b, c);
		}
		else
		{
			scanf("%d%d", &a, &b);
			printf("%I64d\n", QuerynSum(Tree, a, b));
		}
	}
	return 0;
}