POJ3468——A Simple Problem with Integers(线段树区间求和加减)

A Simple Problem with Integers

点我看原题^_^http://poj.org/problem?id=3468

Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 94280 Accepted: 29370
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


解题思路:
    简单线段树区间求和加减题,套模版即可。
    先建树,每个节点为该节点孩子的和,求和只需访问区间相应的节点;但是在区间加减时,直接给该节点+num*(以该节点为根的树的叶子的个数),先不用给叶子加(这就是线段树速度快的原因),当哪天你要去访问该节点的孩子或叶子,再去给它加减。这种题套模版多做题就OK了。


代码实现:
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;
typedef long long LL;
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
const int maxn = 100000+10;
LL sum[maxn<<2], add[maxn<<2];
int N, Q;

void putup(int rt)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}

void addup( int rt, int d )
{
    if( add[rt] )
    {
        add[rt<<1] += add[rt];
        add[rt<<1|1] += add[rt];
        sum[rt<<1] += add[rt] * (d - (d >> 1));
        sum[rt<<1|1] += add[rt] * (d >> 1);
        add[rt] = 0;
    }
}

void build( int l, int r, int rt )
{
    add[rt] = 0;
    if( l==r )
    {
        scanf("%lld",&sum[rt]);
        return ;
    }
    int m = (l+r)>>1;
    build(lson);
    build(rson);
    putup(rt);
}

LL query(int L, int R, int l, int r, int rt)
{
    if( L <= l && R >= r )
    {
        return sum[rt];
    }
    addup(rt, r - l + 1);                       // 检查该节点是否加减过且孩子有没有加减;
    int m = (l+r)>>1;
    LL ret = 0;
    if( L <= m )
        ret += query(L, R, lson);
    if( m < R)
        ret += query(L, R, rson);
    return ret;
}

void update(int L, int R, int num, int l, int r, int rt)
{
    if( L<=l && R>=r )                                   // 只给该节点加减,不必考虑孩子;
    {
        add[rt] += num;
        sum[rt] += (LL)num * (r - l + 1);
        return ;
    }
    addup( rt, r - l + 1);
    int m = (l+r)>>1;
    if( L <= m )
        update(L, R, num, lson);
    if( m < R )
        update(L, R, num, rson);
    putup(rt);
}

int main()
{
    scanf("%d%d",&N,&Q);
    memset(add, 0, sizeof(add));
    memset(sum, 0, sizeof(sum));
    build( 1, N, 1 );
    while( Q-- )
    {
        char op[2];
        int a, b, c;
        scanf("%s",op);
        if( 'Q' == op[0] )
        {
            scanf("%d%d",&a,&b);
            printf("%lld\n",query( a, b, 1, N, 1 ));
        }
        else
        {
            scanf("%d%d%d",&a,&b,&c);
            update(a, b, c, 1, N, 1);
        }
    }
    return 0;
}

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