12299 RMQ with Shifts

最新推荐文章于 2023-01-15 18:33:38 发布

最新推荐文章于 2023-01-15 18:33:38 发布 · 351 阅读

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本文介绍了一种高效的数据结构——线段树，详细讲解了如何使用线段树进行单点更新和区间查询操作，并通过具体代码实现展示了其在解决实际问题中的应用。

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线段树单点更新区间查询

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
const int maxn = 100000 + 7;
const int INF = ~0U >> 1;
struct node {
    int left, right;
    int val;
}str[maxn*5];
int n, q;
int a[maxn];

int BuildTree(int root, int left, int right) {
    str[root].left = left;
    str[root].right = right;
    if(left == right) {
        str[root].val = a[left];
        return a[left];
    }
    int mid = (left + right) / 2;
    int x = BuildTree(root*2, left, mid);
    int y = BuildTree(root*2+1, mid + 1, right);
    return str[root].val = min(x, y);
}

int Update(int root, int id, int val) {
    if(str[root].left == str[root].right && str[root].left == id) {
        a[id] = val;
        str[root].val = val;
        return val;
    }
    if(str[root].left > id || str[root].right < id) return str[root].val;
    int mid = (str[root].left + str[root].right) / 2;
    int x = Update(root*2, id, val);
    int y = Update(root*2+1, id, val);
    return str[root].val = min(x, y);
}

int Query(int root, int left, int right) {
    if(str[root].left >= left && str[root].right  <= right) return str[root].val;
    if(str[root].left > right || str[root].right < left) return INF;
    int x = Query(root*2, left, right);
    int y = Query(root*2+1, left, right);
    return min(x, y);
}

int main() {
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; ++i)
        scanf("%d", &a[i]);
    BuildTree(1, 1, n);
    char que[37] = {0};
    for(int i = 0; i < q; ++i) {
        scanf("%s", que);
        if(que[0] == 'q') {
            int e = 0;
            while(que[e] < '0' || que[e] > '9') ++e;
            int x = atoi(que + e);
            while(que[e] >= '0' && que[e] <= '9') ++e;
            int y = atoi(que + e + 1);
            printf("%d\n", Query(1, x, y));
        } else {
            int b[37] = {0}, k = 0, e = 0;
            while(que[e]) {
                if(que[e] == ')') break;
                while(que[e] < '0' || que[e] > '9') ++e;
                int num = atoi(que + e);
                b[k++] = num;
                while(que[e] >= '0' && que[e] <= '9') ++e;
            }
            int t = b[0], u = a[t];
            for(int i = 0; i < k - 1; ++i)
                Update(1, b[i], a[b[i+1]]);
            Update(1, b[k-1], u);
        }
    }
    return 0;
}