[模板]LCS&LIS

LIS与LCS是经典的动态规划问题.

LIS

AT2827 LIS

假设有序列s:

dp[i]表示以s[i]为结尾的上升子序列的最大长度.O(N²),TLE

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int n, s[100010], dp[100010], ans;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", s + i);

    for (int i = 1; i <= n; i++) {
        dp[i] = 1;
        for (int j = 1; j < i; j++)
            if (s[j] < s[i]) dp[i] = max(dp[j] + 1, dp[i]);
        ans = max(ans, dp[i]);
    }

    printf("%d
", ans);

    return 0;
}

dp[i]表示长度为i的上升子序列末尾元素的最小值.O(NlogN)

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int n, s[100010], dp[100010], len = 1;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", s + i);
        dp[i] = 0x7fffffff;
    }
    dp[1] = s[1];

    for (int i = 2; i <= n; i++)
        if (s[i] > dp[len])
            dp[++len] = s[i];
        else {  
            // 这里当然也可以用upper_bound:
            // int x = upper_bound(dp + 1, dp + len, s[i]) - dp;
            // dp[x] = min(s[i], dp[i]);
            int l = 0, r = len;
            while (l < r) {
                int mid = l + r >> 1;
                if (dp[mid] > s[i])
                    r = mid;
                else
                    l = mid + 1;
            }
            dp[l] = min(s[i], dp[l]);
        }

    printf("%d
", len);

    return 0;
}

AT2827 二分

---

LCS

AcWing最长公共子序列 弱数据,序列为字母

dp[i][j]表示字符串A的i长度前缀与字符串B的j长度前缀的最长公共子序列.

#include <algorithm>
#include <iostream>
using namespace std;

int dp[1001][1001];
char a1[2001], a2[2001];
int n, m;

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> a1[i];
    for (int i = 1; i <= m; i++) cin >> a2[i];

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++) {
            dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            if (a1[i] == a2[j]) dp[i][j] = dp[i - 1][j - 1] + 1;
        }

    cout << dp[n][m];
}

P1439 【模板】最长公共子序列 强数据,序列为全排列.

LCS的O(NlogN)求解依赖于LIS方法.

https://www.luogu.com.cn/blog/blue/solution-p1439

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int n, dp[100010], s1[100010], s2[100010], ans = 1;
int rk[100010];

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", s1 + i);
        rk[s1[i]] = i;
        dp[i] = 0x7FFFFFFF;
    }
    for (int i = 1; i <= n; i++) scanf("%d", s2 + i);
    dp[1] = rk[s2[1]];

    for (int i = 2; i <= n; i++)
        if (rk[s2[i]] > dp[ans])
            dp[++ans] = rk[s2[i]];
        else {
            int x = upper_bound(dp + 1, dp + ans, rk[s2[i]]) - dp;
            dp[x] = min(rk[s2[i]], dp[i]);
            // int l = 0, r = ans;
            // while (l < r) {
            //     int mid = l + r >> 1;
            //     if (dp[mid] > rk[s2[i]])
            //         r = mid;
            //     else
            //         l = mid + 1;
            // }
            // dp[l] = min(rk[s2[i]], dp[l]);
        }

    printf("%d
", ans);

    return 0;
}