dls的数据结构_习题1

将(i, a[i])当作点，对a[i]这个维度扫描，在i这个维度上面建立扫描线
#include<bits/stdc++.h>

using namespace std;
const int N = 2e5+10;

array<int, 4> que[N];
array<int, 2> a[N];
// 线段树维护树链剖剖分的dfs序
struct Node{
    int l, r;
    int minn;
}tr[4*N];


void pushup(Node &F, Node L, Node R){
    F.minn = min(L.minn, R.minn);
}

void pushup(int u){
    pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}

void build(int u, int l, int r){
    if(l == r){
        // 这里注意要加个id,获得序列上l位置的点的是那个点，然后获得他的权值
        tr[u] = {l, r, 0x3f3f3f3f};
    }
    else{
        tr[u] = {l, r, 0x3f3f3f3f};
        int mid = l + r >> 1;
        build(u<<1, l, mid), build(u<<1|1, mid+1, r);
        pushup(u);
    }
}

void modify(int u, int x, int v){
    
    if(tr[u].l >= x && tr[u].r <= x) {
        tr[u].minn = v;
    }
    else{
        int mid = tr[u].l + tr[u].r >> 1;
        if(x <= mid) modify(u<<1, x, v);
        else if(x > mid) modify(u<<1|1, x, v);
        pushup(u);
    }
}

Node query(int u, int l, int r){
    if(tr[u].l >= l && tr[u].r <= r) return tr[u];
    else{
        int mid = tr[u].l + tr[u].r >> 1;
        if(l > mid) return query(u<<1|1, l, r);
        else if(r <= mid) return query(u<<1, l, r);
        else{
            Node res;
            Node L = query(u << 1, l, r), R = query(u << 1 | 1, l, r);
            pushup(res, L, R);
            return res;
        }
    }
}

int res[N];
int main(){
    int n, m; cin >> n >> m;
    memset(res, 0x3f, sizeof res);
    for(int i = 1; i <= n; i ++){
        int x; scanf("%d", &x);
        a[i] = {x, i};
    }
    for(int i = 1; i <= m; i ++){
        int l, r, x; scanf("%d %d %d", &l, &r, &x);
        que[i] = {x, l, r, i};
    }
    sort(a + 1, a + 1 + n);
    reverse(a + 1, a + 1 + n);
    sort(que + 1, que + 1 + m);
    reverse(que + 1, que + 1 + m);
    build(1, 1, n);
    int i, j;
    for(i = 1, j = 1; i <= m + 1; i ++){
        while(j <= n && que[j][0] > a[i][0]){
            res[que[j][3]]  = query(1, que[j][1], que[j][2]).minn;
            j ++;
        }        
        if(i == m + 1) break;
        modify(1, a[i][1], a[i][0]);
    }  

    for(int i = 1; i <= m; i ++){
        if(res[i] != 0x3f3f3f3f) printf("%d\n", res[i]);
        else printf("-1\n");
    }

}

标签：Node,minn,int,res,tr,dls,que,习题,数据结构
来源： https://www.cnblogs.com/njw1123/p/16198520.html