HZNUACM寒假集训Day5小结 线段树 树状数组

线段树

  什么时候用线段树   

  1.统计量可合并

  2.修改量可合并

  3.通过统计量可直接修改统计量

  一句话：满足区间加法即可使用线段树维护信息

  理解Lazy Tage 

  蓝色是要把信息及时维护的节点，红色是本次区间修改操作Lazy Tage下传停止的位置。

  

  模板  Just a Hook  HDU-1698

  

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;

const int maxn = 100050 * 4;  //线段树范围开4倍

struct Tree {
    int l, r, sum, maxx;
};

Tree node[maxn];
int a[maxn];
int lazy[maxn];

void Pushdown(int rt, int m) {
    if (lazy[rt]) {
        lazy[rt << 1] = lazy[rt];
        lazy[rt << 1 | 1] = lazy[rt];
        node[rt << 1].sum = lazy[rt] * (m - (m >> 1));
        node[rt << 1 | 1].sum = lazy[rt] * (m >> 1);
        lazy[rt] = 0;
    }
}

void Pushup(int i) {
    node[i].sum = node[i << 1].sum + node[i << 1 | 1].sum;
    node[i].maxx = max(node[i << 1].maxx, node[i << 1 | 1].maxx);
}

void Build(int i, int l, int r) {
    lazy[i] = 0;
    node[i].l = l;
    node[i].r = r;
    if (l == r) {
        node[i].maxx = a[l];
        node[i].sum = a[l];
        return;
    }
    int mid = l + r >> 1;
    Build(i << 1, l, mid);
    Build(i << 1 | 1, mid + 1, r);
    Pushup(i);
}

int getsum(int i, int l, int r) {
    if (node[i].l == l && node[i].r == r) return node[i].sum;
    int mid = node[i].l + node[i].r >> 1;
    Pushdown(i, node[i].r - node[i].l + 1);
    if (r <= mid) return getsum(i << 1, l, r);
    else if (l > mid) return getsum(i << 1 | 1, l, r);
    else return getsum(i << 1, l, mid) + getsum(i << 1 | 1, mid + 1, r);
}

void update(int i, int l, int r,int v) {
    if (node[i].r == r && node[i].l == l) {
        lazy[i] = v;
        node[i].sum = v * (r - l + 1);
        return;
    }
    if(node[i].l == node[i].r) return;
    int mid = node[i].l + node[i].r >> 1;
    Pushdown(i, node[i].r - node[i].l + 1);
    if (r <= mid) update(i << 1, l, r, v);
    else if (l > mid) update(i << 1 | 1, l, r, v);
    else {
        update(i << 1, l, mid, v);
        update(i << 1 | 1, mid + 1, r, v);
    }
    Pushup(i);
}

int main() {
    int kase = 1;
    int m, n, x, y, z, q;
    string op;
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d", &n, &q);
        for (int i = 1; i <= n; i++) a[i] = 1;
        Build(1, 1, n);
        while (q--) {
            scanf("%d%d%d", &x, &y, &z);
            update(1, x, y, z);
        }
        printf("Case %d:The total value of the hook is %d.\n", kase, getsum(1, 1, n));
        kase++;
    }
    return 0;
}

   “树状数组支持的操作：1、区间和、区间异或和、区间乘积和RMQRMQ（显然，支持的操作都具有交换律，这也算是树状数组的一大特性吧）2、单点修改 ”

   

      code

   

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<cmath>
const double PI = acos(-1.0);
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;

int n, m, tree[2000010];

inline int lowbit(int k) {   //lowbit(x)是x的二进制表达式中最低位的1所对应的值
    return k & -k;
}

void add(int x, int k) {
    while (x <= n) {
        tree[x] += k;
        x += lowbit(x);
    }
}

int sum(int x) {
    int ans = 0;
    while (x) {
        ans += tree[x];
        x -= lowbit(x);
    }
    return ans;
}


int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        int a;
        scanf("%d", &a);
        add(i, a);
    }
    for (int i = 1; i <= m; i++) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        if (a == 1) add(b, c);
        if (a == 2) printf("%d", sum(c) - sum(b - 1));
    }
    return 0;
}