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最新推荐文章于 2018-04-11 22:18:00 发布

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本文介绍了使用线段树进行单点更新加区间求最小值和区间更新加区间求和的两种典型应用案例。通过具体的代码实现展示了如何高效地解决这类问题，并提供了完整的C++实现代码。

单点更新＋区间求最小值

代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define inf 1<<30
#define lchild rt << 1, l, m
#define rchild rt << 1 | 1, m + 1, r
const int maxn = 1e6+100;
int tree[maxn<<2];
int N;

void push_up(int rt)
{
    tree[rt] = min(tree[rt << 1], tree[rt << 1 | 1]);
}

void build(int rt=1, int l=1, int r=N)
{
    if (l == r)
    {
        cin >> tree[rt];
        return;
    }
    int m = (l + r) >> 1;
    build(lchild);
    build(rchild);
    push_up(rt);
}

void update(int p, int delta, int rt=1, int l=1, int r=N)
{
    if (l == r)
    {
        tree[rt] = delta;
        return;
    }
    int m = (l + r) >> 1;
    if (p <= m)
        update(p, delta, lchild);
    else
        update(p, delta, rchild);
    push_up(rt);
}

int query(int L, int R, int rt=1,int l=1, int r=N)
{
    if (L <= l && r <= R) return tree[rt];
    int m = (l + r) >> 1, ret = inf;
    if (L <= m) ret=min(ret,query(L, R, lchild));
    if (R > m)  ret=min(ret,query(L, R, rchild));
    return ret;
}

int main()
{
    scanf("%d",&N);
    build();
    int q;
    scanf("%d",&q);
    while(q--)
    {
        int num;
        scanf("%d",&num);
        if(num==0)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            printf("%d\n",query(l,r));
        }
        else
        {
            int pos,x;
            scanf("%d%d",&pos,&x);
            update(pos, x);
        }
    }
}

区间更新＋区间求和

代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define inf 1<<30
#define lchild rt << 1, l, m
#define rchild rt << 1 | 1, m + 1, r
const int maxn = 1e6+100;
int tree[maxn<<2];
int lazy[maxn<<2];
int N;

void push_up(int rt)
{
    tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}

void push_down(int rt, int len)
{
    tree[rt << 1] = lazy[rt] * (len - (len >> 1));
    lazy[rt << 1] = lazy[rt];
    tree[rt << 1 | 1] = lazy[rt] * (len >> 1);
    lazy[rt << 1 | 1] = lazy[rt];
    lazy[rt] = 0;
}

void build(int rt=1, int l=1, int r=N)
{
    lazy[rt]=0;
    if (l == r)
    {
        cin >> tree[rt];
        return;
    }
    int m = (l + r) >> 1;
    build(lchild);
    build(rchild);
    push_up(rt);
}

void update(int L, int R, int delta, int rt = 1, int l = 1, int r = N) {
    if (L <= l && r <= R) {
        tree[rt] = delta * (r - l + 1);
        lazy[rt] = delta;
        return;
    }
    if (lazy[rt]) push_down(rt, r - l + 1);
    int m = (l + r) >> 1;
    if (L <= m) update(L, R, delta, lchild);
    if (R > m)  update(L, R, delta, rchild);
    push_up(rt);
}

int query(int L, int R, int rt = 1, int l = 1, int r = N) {
    if (L <= l && r <= R) return tree[rt];
    if (lazy[rt]) push_down(rt, r - l + 1);
    int m = (l + r) >> 1, ret = 0;
    if (L <= m) ret += query(L, R, lchild);
    if (R > m)  ret += query(L, R, rchild);
    return ret;
}

int main()
{
    scanf("%d",&N);
    build();
    int q;
    scanf("%d",&q);
    while(q--)
    {
        int num;
        scanf("%d",&num);
        if(num==0)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            printf("%d\n",query(l,r));
        }
        else
        {
            int l,r,x;
            scanf("%d%d%d",&l,&r,&x);
            update(l,r,x);
        }
    }
}