题解：P11876 收徒！

#include<bits/stdc++.h>
using namespace std;
const int N = 5e3+5;
int h, w, n, x, y;
int dp[N][N], mp[N][N];
char pre[N][N];
void dfs(int x, int y) {
	if (x == 1 && y == 1) return ;
	if (pre[x][y] == 'D')	dfs(x - 1, y);
	else	dfs(x, y - 1);
	cout << pre[x][y];
}
int main() {
	cin >> h >> w >> n;
	for (int i = 1; i <= n; i++) {
		cin >> x >> y;
		mp[x][y] = 1;
	}
	for (int i = 2; i <= n; i++)
		dp[i][1] = dp[i - 1][1] + mp[i][1], pre[i][1] = 'D';
	for (int j = 2; j <= w; j++)
		dp[1][j] = dp[1][j - 1] + mp[1][j], pre[1][j] = 'R';
	for (int i = 2; i <= h; i++)
		for (int j = 2; j <= w; j++) {
			if (dp[i - 1][j] > dp[i][j - 1]) {
				pre[i][j] = 'D';
				dp[i][j] = dp[i - 1][j] + mp[i][j];
			} else {
				pre[i][j] = 'R';
				dp[i][j] = dp[i][j - 1] + mp[i][j];
			}
		}
	cout << dp[h][w] << endl;
	dfs(h, w);
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int h, w, n, c[N], dp[N], nowx[N], nowy[N], k;
int pre[N];		//记录第i个数的前一项的下标
int lst[N];		//记录长度为len的最后一项的下标
string ans;
struct node {
	int x, y;
	bool operator<(node &b) {
		if (x == b.x) return y < b.y;
		return x < b.x;
	}
} a[N];
void dfs(int now) {
	if (now == 0) return;
	dfs(pre[now]);
	++k;
	nowx[k] = a[now].x, nowy[k] = a[now].y;
}
int main() {
	cin >> h >> w >> n;
	for (int i = 1; i <= n; i++) cin >> a[i].x >> a[i].y;
	sort(a + 1, a + 1 + n);
	dp[1] = a[1].y;
	lst[1] = 1, pre[1] = 0;
	int len = 1;
	for (int i = 2; i <= n; i++) {
		if (dp[len] <= a[i].y) {		//如果这一项能接在最长的一项后面
			pre[i] = lst[len];			//当前项的前驱为之前长度为len的最小的最长项
			dp[++len] = a[i].y;			//记录当前长度为len的子序列的最后一项
			lst[len] = i;				//记录当前长度为len的最后一项的下标
		} else {
			int x = upper_bound(dp + 1, dp + 1 + len, a[i].y) - dp;	//查找等于len的
			pre[i] = lst[x - 1];
			dp[x] = a[i].y;
			lst[x] = i;
		}
	}
	cout << len << endl;
	nowx[1] = nowy[1] = k = 1;
	dfs(lst[len]);						//递归记录路径
	++k, nowx[k] = h, nowy[k] = w;
	for (int i = 2; i <= k; i++) {		//根据曼哈顿距离构造答案
		string D(nowx[i] - nowx[i - 1], 'D');
		string R(nowy[i] - nowy[i - 1], 'R');
		ans += D + R;
//		cout << nowx[i] << " " << nowy[i] << endl;
	}
	cout << ans;
	return 0;
}