POJ3486 线段树

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2017年3月15日 | ljfcnyali
POJ3486是一道经典线段树题目，题目大意是：
给出了一个序列，你需要处理如下两种询问。
“C a b c”表示给[a, b]区间中的值全部增加c (-10000 ≤ c ≤ 10000)。
“Q a b” 询问[a, b]区间中所有值的和。
Input
第一行包含两个整数N, Q。1 ≤ N,Q ≤ 100000.
第二行包含n个整数，表示初始的序列A (-1000000000 ≤ Ai ≤ 1000000000)。
接下来Q行询问。
Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

4559151234455915

题目分析：
线段树，并且单纯的线段树会超时，因为在将a到b的点全部加上c时，步骤太多，会超时，需要优化–即lazy算法；
Lazy算法：
在将a~b点全部加c时，不要加到每个点，在表示区间的root结构体上增加一个inc域，将要加的值赋给这个inc域，然后就不要再往下了。
在求区间和时，将root中的inc值赋给要求的区间，并且将该节点的root置零。

/*************************************************************************    > File Name: POJ3468.cpp    > Author: ljf-cnyali    > Mail: ljfcnyali@gmail.com     > 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;}

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