1348F - Phoenix and Memory

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求一个全排列，并且$a[i]\leq ans[i]\leq b[i]$。 保证有解，判断是否存在多组解，输出两组。

考虑生成一组解。考虑一定存在，那么贪心的考虑 对于$a[i]\leq x$,选择比较小的的$b[i]\geq x$,因为之后大的选择权利之后大，优先把选择权利小的让出去。（因为有解这个$b[i]$一定存在)。用一个$set$维护就很方便。

$p_i$表示$i$出现的位置。 现在枚举每个位置，假设$j$位置可以交换，$a[i]\leq ans[j]\leq b[i]$,$a[j]\leq ans[i]\leq b[j]$。假设$ans[j]>ans[i]\rightarrow a[j]\leq ans[i]<ans[j]\leq b[i]$。

也就是查询$min(ans[i],b[i])\leq ans[i]$ ,线段树解决。

代码 
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#include <bits/stdc++.h>
#include <set>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 2e5 + 10;
const int mod = 1e9 + 7;

int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}
int p[N];
int ans[N];
const int SEN = 2e5 + 10;
int a[N], b[N];
struct segmentTree
{

    pii sum[SEN << 2];

    void pushup(int pos)
    {
        sum[pos] = min(sum[pos << 1 | 1], sum[pos << 1]);
    }
    pii query(int ql, int qr, int pos, int l, int r)
    {
        if (ql <= l && r <= qr)
        {
            return sum[pos];
        }
        //pushdown(pos, l, r);
        int mid = (l + r) >> 1;
        pii ans = mk(1e9, 0);
        if (ql <= mid)
            ans = min(ans, query(ql, qr, pos << 1, l, mid));
        if (qr > mid)
            ans = min(ans, query(ql, qr, pos << 1 | 1, mid + 1, r));
        return ans;
    }
    void build(int pos, int l, int r)
    {
        sum[pos].first = 1e9;
        if (l == r)
        {
            sum[pos].first = a[p[l]];
            sum[pos].second = l;
            return;
        }
        int mid = (l + r) >> 1;
        build(pos << 1, l, mid);
        build(pos << 1 | 1, mid + 1, r);
        pushup(pos);
    }
} T;

vector<pii> l[N];

int main()
{
    int n = read();
    for (int i = 1; i <= n; i++)
    {
        a[i] = read(), b[i] = read();
        l[a[i]].push_back(mk(b[i], i));
    }
    set<pii> s;
    for (int i = 1; i <= n; i++)
    {
        s.insert(l[i].begin(), l[i].end());
        ans[s.begin()->second] = i;
        p[i] = s.begin()->second;
        s.erase(s.begin());
    }
    T.build(1, 1, n);
    for (int i = 1; i <= n; i++)
    {
        pii k = T.query(ans[i] + 1, b[i], 1, 1, n);
        //cout << k.first << " " << k.second << endl;

        if (k.first <= ans[i])
        {

            printf("NO\n");
            for (int j = 1; j <= n; j++)
                printf("%d ", ans[j]);
            puts("");
            swap(ans[i], ans[p[k.second]]);
            for (int j = 1; j <= n; j++)
                printf("%d ", ans[j]);
            return 0;
        }
    }
    printf("YES\n");
    for (int j = 1; j <= n; j++)
        printf("%d ", ans[j]);
    puts("");
}