HDU 1423 Greatest Common Increasing Subsequence LCIS

题目链接：

题目

Greatest Common Increasing Subsequence
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)

问题描述

This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

输入

Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.

输出

output print L - the length of the greatest common increasing subsequence of both sequences.

样例

input
1
5
1 4 2 5 -12
4
-12 1 2 4

output
2

题意

求两个串的最长公共上升子序列（LCIS）

题解

dp[j]表示第一个串的前i个和第二个串的前j个的以b[j]结尾的公共最长上升子序列的长度。

代码

O(n^3):

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

const int maxn = 555;

int dp[maxn];
int a[maxn], b[maxn];
int n, m;

void init() {
    memset(dp, 0, sizeof(dp));
}

int main() {
    int tc;
    scanf("%d", &tc);
    while (tc--) {
        init();
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        scanf("%d", &m);
        for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
        int ans = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                if (a[i] == b[j]) {
                    dp[j] = 1;//b[0]是非法的！
                    for (int k = 0; k < j; k++) {
                        if (b[k] < b[j] && dp[j] < dp[k] + 1) {
                            dp[j] = dp[k] + 1;
                        }
                    }
                }
                ans = max(ans, dp[j]);
            }
        }
        printf("%d\n", ans);
        if (tc) printf("\n");
    }
    return 0;
}

O(n^2):k循环其实可以不用，用Max一边扫j，一边记录就可以了。

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

const int maxn = 555;

int dp[maxn];
int a[maxn], b[maxn];
int n, m;

void init() {
    memset(dp, 0, sizeof(dp));
}

int main() {
    int tc;
    scanf("%d", &tc);
    while (tc--) {
        init();
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        scanf("%d", &m);
        for (int i = 1; i <= m; i++) scanf("%d", &b[i]);
        int ans = 0;
        for (int i = 1; i <= n; i++) {
            int Max = 0;
            for (int j = 1; j <= m; j++) {
                if (b[j]<a[i]&&Max<dp[j]) {
                    Max = dp[j];
                }
                else if (a[i] == b[j]) {
                    dp[j] = Max + 1;
                }
                ans = max(ans, dp[j]);
            }
        }
        printf("%d\n", ans);
        if (tc) printf("\n");
    }
    return 0;
}

来源：https://www.cnblogs.com/fenice/p/5664863.html