HDU4512完美队形I && HDU1423 Greatest Common Increasing Subsequence (LCIS)

填坑的时候又到啦，校赛因为不会LCIS所以吃了大亏，这里要补起来。LCIS就是在两个串里找最长上升子序列，相关的博客有很多，这里自己就不写那么多了。

http://www.cnblogs.com/jackge/archive/2013/05/16/3081793.html

http://www.cnblogs.com/gj-Acit/p/3236384.html

上面两个博客对于O(n^2)的做法讲解的比较详细，大家可以参考一下。

贴两记代码

HDU1423

#pragma warning(disable:4996)
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>
#include <cstdio>
using namespace std;

#define maxn 550
int a[maxn];
int b[maxn];
int n1, n2;;
int dp[maxn][maxn];

int main()
{
    int T; cin >> T;
    while (T--){
        cin >> n1;
        for (int i = 1; i <= n1; i++) scanf("%d", a + i);
        cin >> n2;
        for (int i = 1; i <= n2; i++) scanf("%d", b + i);
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n1; i++){
            int tmp = 0;
            for (int j = 1; j <= n2; j++){
                dp[i][j] = dp[i - 1][j];
                if (a[i] > b[j] && dp[i - 1][j] > tmp) tmp = dp[i - 1][j];
                if (a[i] == b[j]) dp[i][j] = tmp + 1;
            }
        }
        int ans = 0;
        for (int i = 1; i <= n1; i++){
            ans = max(ans, dp[n1][i]);
        }
        printf("%d\n", ans);
        if (T) puts("");
    }
    return 0;
}

HDU4512

#pragma warning(disable:4996)
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <algorithm>
#include <cstdio>
using namespace std;

#define maxn 550

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

int main()
{
    int T; cin >> T;
    while (T--)
    {
        cin >> n;
        for (int i = 1; i <= n; i++){
            scanf("%d", a + i);
            b[n + 1 - i] = a[i];
        }
        int ans = 0;
        memset(dp, 0, sizeof(dp));
        for (int i = 1; i <= n; i++){
            int tmp = 0;
            for (int j = 1; j <= n+1-i; j++){
                dp[i][j] = dp[i - 1][j];
                if (a[i] > b[j] && dp[i - 1][j] > tmp) tmp = dp[i - 1][j];
                if (a[i] == b[j]){
                    dp[i][j] = max(dp[i][j],tmp + 1);
                }
                if (i < n + 1 - j) ans = max(ans, dp[i][j] * 2);
                else ans = max(ans, dp[i][j] * 2 - 1);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}