Codeforces 622C Not Equal on a Segment 【线段树 or dp】

最新推荐文章于 2019-08-09 15:29:24 发布

笑着走完自己的路

最新推荐文章于 2019-08-09 15:29:24 发布

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分类专栏： codeforces 线段树我就是DP

本文链接：https://blog.youkuaiyun.com/chenzhenyu123456/article/details/50659399

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本文介绍了一种使用线段树解决特定查询问题的方法，即在给定区间内找到第一个不等于给定值的元素的索引。通过维护区间的最大值、最小值及其位置，可以在对数时间内高效地解决该问题。

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You are given array a with n integers and m queries. The i-th query is given with three integers l_i, r_i, x_i.

For the i-th query find any position p_i (l_i ≤ p_i ≤ r_i) so that a_{p_i} ≠ x_i.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·10⁵) — the number of elements in a and the number of queries.

The second line contains n integers a_i (1 ≤ a_i ≤ 10⁶) — the elements of the array a.

Each of the next m lines contains three integers l_i, r_i, x_i (1 ≤ l_i ≤ r_i ≤ n, 1 ≤ x_i ≤ 10⁶) — the parameters of the i-th query.

Output

Print m lines. On the i-th line print integer p_i — the position of any number not equal to x_i in segment [l_i, r_i] or the value - 1 if there is no such number.

Sample test(s)

input
6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2

output
2
6
-1
4

题意：给你n个数和m次查询，问你[l, r] 里面!= x的数下标，任意一个都可以。

我的思路不是标程。。。

思路：用一棵线段树维护区间最大值Max，最小值Min，以及对应的位置pa，pi。发现无解的情况只有Max == Min == x，其余情况均有解，分类讨论就好了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 1000000
#define eps 1e-8
#define MAXN (1000000+10)
#define MAXM (2000000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
struct Tree{
    int l, r, Max, Min, pa, pi;
};
Tree tree[MAXN<<2];
void PushUp(int o)
{
    tree[o].Min = min(tree[ll].Min, tree[rr].Min);
    tree[o].Max = max(tree[ll].Max, tree[rr].Max);
    if(tree[o].Min == tree[ll].Min) tree[o].pi = tree[ll].pi;
    else tree[o].pi = tree[rr].pi;
    if(tree[o].Max == tree[ll].Max) tree[o].pa = tree[ll].pa;
    else tree[o].pa = tree[rr].pa;
}
void Build(int o, int l, int r)
{
    tree[o].l = l; tree[o].r = r;
    if(l == r)
    {
        int a; Ri(a);
        tree[o].Min = tree[o].Max = a;
        tree[o].pa = tree[o].pi = l;
        return ;
    }
    int mid = (l + r) >> 1;
    Build(lson); Build(rson);
    PushUp(o);
}
int Query(int o, int L, int R, int op)
{
    if(L == tree[o].l && R == tree[o].r)
    {
        if(op == 1) return tree[o].Max;
        else if(op == 2) return tree[o].Min;
        else if(op == 3) return tree[o].pa;
        else return tree[o].pi;
    }
    int mid = (tree[o].l + tree[o].r) >> 1;
    if(R <= mid) return Query(ll, L, R, op);
    else if(L > mid) return Query(rr, L, R, op);
    else
    {
        if(op == 1) return max(Query(ll, L, mid, op), Query(rr, mid+1, R, op));
        else if(op == 2) return min(Query(ll, L, mid, op), Query(rr, mid+1, R, op));
        else if(op == 3)
        {
            int ans1 = Query(ll, L, mid, 1), ans2 = Query(rr, mid+1, R, 1);
            int ans = max(ans1, ans2);
            if(ans == ans1) return Query(ll, L, mid, 3);
            else return Query(rr, mid+1, R, 3);
        }
        else
        {
            int ans1 = Query(ll, L, mid, 2), ans2 = Query(rr, mid+1, R, 2);
            int ans = min(ans1, ans2);
            if(ans == ans1) return Query(ll, L, mid, 4);
            else return Query(rr, mid+1, R, 4);
        }
    }
}
int main()
{
    int n, m; Ri(n); Ri(m);
    Build(1, 1, n);
    int Min, Max, pa, pi;
    for(int i = 0; i < m; i++)
    {
        int l, r, x;
        Ri(l); Ri(r); Ri(x);
        Max = Query(1, l, r, 1);
        Min = Query(1, l, r, 2);
        pa = Query(1, l, r, 3);
        pi = Query(1, l, r, 4);
        //printf("%d %d %d %d\n", Max, Min, pa, pi);
        int ans;
        if(Min == Max && Min == x) ans = -1;
        else if(Min != x) ans = pi;
        else if(Max != x) ans = pa;
        Pi(ans);
    }
    return 0;
}

标程解法：设置dp[i]记录最大的j使得(1 <= j < i)a[j] != a[i]。

最后对右边界分类讨论即可。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <string>
#define INF 1000000
#define eps 1e-8
#define MAXN (1000000+10)
#define MAXM (2000000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
typedef pair<int, int> pii;
int a[MAXN];
int dp[MAXN];
int main()
{
    int n, m; Ri(n); Ri(m);
    for(int i = 1; i <= n; i++)
    {
        Ri(a[i]);
        if(i == 1) dp[i] = -1;
        else
        {
            if(a[i] == a[i-1]) dp[i] = dp[i-1];
            else dp[i] = i-1;
        }
    }
    W(m)
    {
        int l, r, x;
        Ri(l); Ri(r); Ri(x);
        if(a[r] == x)
        {
            if(dp[r] < l) Pi(-1);
            else Pi(dp[r]);
        }
        else
            Pi(r);
    }
    return 0;
}