notonlysuccess大神的线段树完全版

在大神的网站进不去的时候可以过来看看，另外道客巴巴有个排版比较好的文档，外观派可以去看看http://www.doc88.com/p-2728103209174.html

很早前写的那篇线段树专辑至今一直是本博客阅读点击量最大的一片文章,当时觉得挺自豪的,还去pku打广告,但是现在我自己都不太好意思去看那篇文章了,觉得当时的代码风格实在是太丑了,很多线段树的初学者可能就是看着这篇文章来练习的,如果不小心被我培养出了这么糟糕的风格,实在是过意不去,正好过几天又要给集训队讲解线段树,所以决定把这些题目重新写一遍,顺便把近年我接触到的一些新题更新上去~;并且学习了splay等更高级的数据结构后对线段树的体会有更深了一层,线段树的写法也就比以前飘逸,简洁且方便多了.

在代码前先介绍一些我的线段树风格:




maxn是题目给的最大区间,而节点数要开4倍,确切的来说节点数要开大于maxn的最小2x的两倍


lson和rson分辨表示结点的左儿子和右儿子,由于每次传参数的时候都固定是这几个变量,所以可以用预定于比较方便的表示


以前的写法是另外开两个个数组记录每个结点所表示的区间,其实这个区间不必保存,一边算一边传下去就行,只需要写函数的时候多两个参数,结合lson和rson的预定义可以很方便


PushUP(int rt)是把当前结点的信息更新到父结点


PushDown(int rt)是把当前结点的信息更新给儿子结点


rt表示当前子树的根(root),也就是当前所在的结点


整理这些题目后我觉得线段树的题目整体上可以分成以下四个部分:




单点更新:最最基础的线段树,只更新叶子节点,然后把信息用PushUP(int r)这个函数更新上来


hdu1166 敌兵布阵
http://acm.hdu.edu.cn/showproblem.php?pid=1166


题意:O(-1)


思路:O(-1)


线段树功能:update:单点增减 query:区间求和


#include <cstdio>


 


#define lson l , m , rt << 1


#define rson m + 1 , r , rt << 1 | 1


const int maxn = 55555;


int sum[maxn<<2];


void PushUP(int rt) {


        sum[rt] = sum[rt<<1] + sum[rt<<1|1];


}


void build(int l,int r,int rt) {


        if (l == r) {


                scanf("%d",&sum[rt]);


                return ;


        }


        int m = (l + r) >> 1;


        build(lson);


        build(rson);


        PushUP(rt);


}


void update(int p,int add,int l,int r,int rt) {


        if (l == r) {


                sum[rt] += add;


                return ;


        }


        int m = (l + r) >> 1;


        if (p <= m) update(p , add , lson);


        else update(p , add , rson);


        PushUP(rt);


}


int query(int L,int R,int l,int r,int rt) {


        if (L <= l && r <= R) {


                return sum[rt];


        }


        int m = (l + r) >> 1;


        int ret = 0;


        if (L <= m) ret += query(L , R , lson);


        if (R > m) ret += query(L , R , rson);


        return ret;


}


int main() {


        int T , n;


        scanf("%d",&T);


        for (int cas = 1 ; cas <= T ; cas ++) {


                printf("Case %d:\n",cas);


                scanf("%d",&n);


                build(1 , n , 1);


                char op[10];


                while (scanf("%s",op)) {


                        if (op[0] == 'E') break;


                        int a , b;


                        scanf("%d%d",&a,&b);


                        if (op[0] == 'Q') printf("%d\n",query(a , b , 1 , n , 1));


                        else if (op[0] == 'S') update(a , -b , 1 , n , 1);


                        else update(a , b , 1 , n , 1);


                }


        }


        return 0;


}






hdu1754 I Hate It


http://acm.hdu.edu.cn/showproblem.php?pid=1754


题意:O(-1)


思路:O(-1)


线段树功能:update:单点替换 query:区间最值


#include <cstdio>


#include <algorithm>


using namespace std;


 


#define lson l , m , rt << 1


#define rson m + 1 , r , rt << 1 | 1


const int maxn = 222222;


int MAX[maxn<<2];


void PushUP(int rt) {


        MAX[rt] = max(MAX[rt<<1] , MAX[rt<<1|1]);


}


void build(int l,int r,int rt) {


        if (l == r) {


                scanf("%d",&MAX[rt]);


                return ;


        }


        int m = (l + r) >> 1;


        build(lson);


        build(rson);


        PushUP(rt);


}


void update(int p,int sc,int l,int r,int rt) {


        if (l == r) {


                MAX[rt] = sc;


                return ;


        }


        int m = (l + r) >> 1;


        if (p <= m) update(p , sc , lson);


        else update(p , sc , rson);


        PushUP(rt);


}


int query(int L,int R,int l,int r,int rt) {


        if (L <= l && r <= R) {


                return MAX[rt];


        }


        int m = (l + r) >> 1;


        int ret = 0;


        if (L <= m) ret = max(ret , query(L , R , lson));


        if (R > m) ret = max(ret , query(L , R , rson));


        return ret;


}


int main() {


        int n , m;


        while (~scanf("%d%d",&n,&m)) {


                build(1 , n , 1);


                while (m --) {


                        char op[2];


                        int a , b;


                        scanf("%s%d%d",op,&a,&b);


                        if (op[0] == 'Q') printf("%d\n",query(a , b , 1 , n , 1));


                        else update(a , b , 1 , n , 1);


                }


        }


        return 0;


}


hdu1394 Minimum Inversion Number


http://acm.hdu.edu.cn/showproblem.php?pid=1394


题意:求Inversion后的最小逆序数


思路:用O(nlogn)复杂度求出最初逆序数后,就可以用O(1)的复杂度分别递推出其他解


线段树功能:update:单点增减 query:区间求和


#include <cstdio>


#include <algorithm>


using namespace std;


 


#define lson l , m , rt << 1


#define rson m + 1 , r , rt << 1 | 1


const int maxn = 5555;


int sum[maxn<<2];


void PushUP(int rt) {


        sum[rt] = sum[rt<<1] + sum[rt<<1|1];


}


void build(int l,int r,int rt) {


        sum[rt] = 0;


        if (l == r) return ;


        int m = (l + r) >> 1;


        build(lson);


        build(rson);


}


void update(int p,int l,int r,int rt) {


        if (l == r) {


                sum[rt] ++;


                return ;


        }


        int m = (l + r) >> 1;


        if (p <= m) update(p , lson);


        else update(p , rson);


        PushUP(rt);


}


int query(int L,int R,int l,int r,int rt) {


        if (L <= l && r <= R) {


                return sum[rt];


        }


        int m = (l + r) >> 1;


        int ret = 0;


        if (L <= m) ret += query(L , R , lson);


        if (R > m) ret += query(L , R , rson);


        return ret;


}


int x[maxn];


int main() {


        int n;


        while (~scanf("%d",&n)) {


                build(0 , n - 1 , 1);


                int sum = 0;


                for (int i = 0 ; i < n ; i ++) {


                        scanf("%d",&x[i]);


                        sum += query(x[i] , n - 1 , 0 , n - 1 , 1);


                        update(x[i] , 0 , n - 1 , 1);


                }


                int ret = sum;


                for (int i = 0 ; i < n ; i ++) {


                        sum += n - x[i] - x[i] - 1;


                        ret = min(ret , sum);


                }


                printf("%d\n",ret);


        }


        return 0;


}


      成段更新


(通常这对初学者来说是一道坎),需要用到延迟标记(或者说懒惰标记),简单来说就是每次更新的时候不要更新到底,用延迟标记使得更新延迟到下次需要更新or询问到的时候


hdu1698 Just a Hook


http://acm.hdu.edu.cn/showproblem.php?pid=1698


题意:O(-1)


思路:O(-1)


线段树功能:update:成段替换 (由于只query一次总区间,所以可以直接输出1结点的信息


#include <cstdio>


#include <algorithm>


using namespace std;


 


#define lson l , m , rt << 1


#define rson m + 1 , r , rt << 1 | 1


const int maxn = 111111;


int h , w , n;


int col[maxn<<2];


int sum[maxn<<2];


void PushUp(int rt) {


        sum[rt] = sum[rt<<1] + sum[rt<<1|1];


}


void PushDown(int rt,int m) {


        if (col[rt]) {


                col[rt<<1] = col[rt<<1|1] = col[rt];


                sum[rt<<1] = (m - (m >> 1)) * col[rt];


                sum[rt<<1|1] = (m >> 1) * col[rt];


                col[rt] = 0;


        }


}


void build(int l,int r,int rt) {


        col[rt] = 0;


        sum[rt] = 1;


        if (l == r) return ;


        int m = (l + r) >> 1;


        build(lson);


        build(rson);


        PushUp(rt);


}


void update(int L,int R,int c,int l,int r,int rt) {


        if (L <= l && r <= R) {


                col[rt] = c;


                sum[rt] = c * (r - l + 1);


                return ;


        }


        PushDown(rt , r - l + 1);


        int m = (l + r) >> 1;


        if (L <= m) update(L , R , c , lson);


        if (R > m) update(L , R , c , rson);


        PushUp(rt);


}


int main() {


        int T , n , m;


        scanf("%d",&T);


        for (int cas = 1 ; cas <= T ; cas ++) {


                scanf("%d%d",&n,&m);


                build(1 , n , 1);


                while (m --) {


                        int a , b , c;


                        scanf("%d%d%d",&a,&b,&c);


                        update(a , b , c , 1 , n , 1);


                }


                printf("Case %d: The total value of the hook is %d.\n",cas , sum[1]);


        }


        return 0;


}