标签：int char content nod return padding else 过程 print
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;
}
标签：int char content nod return padding else 过程 print
原文地址：http://www.cnblogs.com/demian/p/6082915.html
