POJ3486 線段樹

2017年3月15日 | ljfcnyali
POJ3486是一道經典線段樹題目，題目大意是：
給出了一個序列，你需要處理如下兩種詢問。
“C a b c”表示給[a, b]區間中的值全部增加c (-10000 ≤ c ≤ 10000)。
“Q a b” 詢問[a, b]區間中所有值的和。
Input
第一行包含兩個整數N, Q。1 ≤ N,Q ≤ 100000.
第二行包含n個整數，表示初始的序列A (-1000000000 ≤ Ai ≤ 1000000000)。
接下來Q行詢問。
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
1
2
3
4
4
55
9
15  

題目分析：
線段樹，並且單純的線段樹會超時，因爲在將a到b的點全部加上c時，步驟太多，會超時，需要優化–即lazy算法；
Lazy算法：
在將a~b點全部加c時，不要加到每個點，在表示區間的root結構體上增加一個inc域，將要加的值賦給這個inc域，然後就不要再往下了。
在求區間和時，將root中的inc值賦給要求的區間，並且將該節點的root置零。

/*************************************************************************
    > File Name: POJ3468.cpp
    > Author: ljf-cnyali
    > Mail: [email protected] 
    > Created Time: 2017/3/15 19:19:37
 ************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>

using namespace std;

#define REP(i, a, b) for(long long i = (a), _end_ = (b);i <= _end_; ++ i)
#define mem(a) memset((a), 0, sizeof(a))
#define str(a) strlen(a)

long long n, m;

struct Line_Tree {
    long long l, r;
    Line_Tree * lson, * rson;
    long long inc, key;
};

Line_Tree * Build(long long left, Line_Tree * root, long long right) {
    root -> l = left;
    root -> r = right;
    root -> key = 0;
    root -> inc = 0;
    long long mid = (left + right) / 2;
    if(left == right) {
        root -> lson = NULL;
        root -> rson = NULL;
        return root;
    }
    Line_Tree * Rootright;
    Rootright = new Line_Tree;
    root -> rson = Rootright;
    Build(mid + 1, root -> rson, right);
    Line_Tree * Rootleft;
    Rootleft = new Line_Tree;
    root -> lson = Rootleft;
    Build(left, root -> lson, mid);
}

void insert(long long now, Line_Tree * root, long long x) {
    long long mid = (root -> l + root -> r) / 2;
    if(root -> l == now && root -> r == now) {
        root -> key += x;
        return ;
    }
    root -> key += x;
    if(root -> l <= now && now <= mid)
        insert(now, root -> lson, x);
    else 
        insert(now, root -> rson, x);
}

void addsum(long long left, long long right, long long add, Line_Tree * root) {
    if(root -> l == left && root -> r == right) {
        root -> inc += add;
        return ;
    }
    root -> key += add * (right - left + 1);
    long long mid = (root -> r + root -> l) / 2;
    if(right <= mid) 
        addsum(left, right, add, root -> lson);
    else if(left >= mid + 1) 
        addsum(left, right, add, root -> rson);
    else {
        addsum(left, mid, add, root -> lson);
        addsum(mid + 1, right, add, root -> rson);
    }
}

long long findsum(long long left, long long right, Line_Tree * root) {
    if(left == root -> l && right == root -> r) 
        return root -> key + root -> inc * (right - left + 1);  
    root -> key += root -> inc * (right - left + 1);    
    long long mid = (root -> l + root -> r) / 2;
    addsum(root -> l, mid, root -> inc, root -> lson);
    addsum(mid + 1, root -> r, root -> inc, root -> rson);
    root -> inc = 0;
    if(left >= root -> l && right <= mid)
        return findsum(left, right, root -> lson);
    else if(left >= mid + 1 && right <= root -> r)
        return findsum(left, right, root -> rson);
    else 
        return findsum(left, mid, root -> lson) + findsum(mid + 1, right, root -> rson);
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    while(scanf("%lld%lld", &n, &m) != EOF) {
        Line_Tree * Root;
        Root = new Line_Tree;
        Build(1, Root, n);
        long long x;
        REP(i, 1, n) {
            scanf("%lld", &x);
            insert(i, Root, x);
        }
        long long y, z;
        char c;
        REP(i, 1, m) {
            getchar();
            scanf("%c", &c);
            if(c == 'C') {
                scanf("%lld%lld%lld", &x, &y, &z);
                addsum(x, y, z, Root);
            }
            else {
                scanf("%lld%lld", &x, &y);
                printf("%lld\n", findsum(x, y, Root));
            }
        }
    }
    return 0;
}

本文轉自：http://ljf-cnyali.cn/index.php/archives/23