CF4D D. Mysterious Present

做题链接： D. Mysterious Present
题目：

给出一个限制(w, h),和n个物品的二维信息（wi，hi）求物品二维都满足wi > w, hi > h的前提下的最长二维严格上升子序列以及长度(wi > wi - 1, hi > hi - 1)
如果找不到任何物品就直接输出0.
物品的顺序可以任意，不做要求
1≤n≤5000， 1≤w,h≤1,000,000

解决：

n只有5000，显然本题是具有二维限制的求最长上升子序问题，并输出子序
先按照(x,y)为第一、第二关键字升序排序，然后做一遍最长上升子序，并且注意记录状态转移的路径。

代码：

#include <bits/stdc++.h>
using namespace std;
const int N = 5010;
struct Node{int x, y;int id;
};
bool cmp(Node a, Node b)
{if (a.x == b.x) return a.y < b.y;return a.x < b.x;
}
int n, xx, yy;
vector<Node> q;
struct {int val, pre;
}dp[N];int main()
{cin >> n >> xx >> yy;for (int i = 1; i <= n; i ++ ){int x, y; scanf("%d%d", &x, &y);if (x <= xx || y <= yy) continue;q.push_back({x, y, i});}q.push_back({xx, yy, -1});sort(q.begin(), q.end(), cmp);int mx = -1, id = -1;for (int i = 1; i < q.size(); i ++ )for (int j = 0; j < i && q[j].x < q[i].x; j ++ )if (q[j].y < q[i].y){if (dp[i].val < dp[j].val + 1){dp[i].val = dp[j].val + 1;dp[i].pre = j;if (mx < dp[i].val){mx = dp[i].val;id = i;}}}if (q.size() == 1){cout << 0 << endl;return 0;}// cout << mx << " " << id << endl;// cout << id << " " << q[id].id  <<endl;vector<int> ans;while (id != -1 && id != 0){ans.push_back(q[id].id);// printf("now.id = %d, pre.id = %d\n", q[id].id, q[dp[id].pre].id);id = dp[id].pre;}cout << ans.size() << endl;reverse(ans.begin(), ans.end());for (int x : ans) cout << x << " ";cout << endl;return 0;
}

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