HDU 5904 LCIS DP

http://acm.hdu.edu.cn/showproblem.php?pid=5904

给定两个序列，要求算出其最长的公共上升子序列，

并且这个子序列公差是1.

考虑dp

dpa[val]表示在a数组中，以val这个数字结尾，最长上升1的子序列。dpa[val] = dpa[val - 1] + 1

同理dpb[val]

然后，对于每个相同的数字，需要取min(dpa[val], dpb[val])即可。

如果dpa[5] = 3是这个序列，3 4 5

然后dpb[5] = 5是这个序列 1 2 3 4 5

那么公共部分就是他们的最小值。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>

const int maxn = 100000 + 20;
int a[maxn];
int b[maxn];
int dpa[maxn];
int dpb[maxn];
void work() {
    memset(dpa, 0, sizeof dpa);
    memset(dpb, 0, sizeof dpb);
    int n, m;
    int mx = -inf;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
        mx = max(mx, a[i]);
    }
    int mx2 = -inf;
    for (int i = 1; i <= m; ++i) {
        scanf("%d", &b[i]);
        mx2 = max(mx2, b[i]);
    }
    mx = min(mx, mx2);
    dpa[a[1]] = 1;
    for (int i = 2; i <= n; ++i) {
        dpa[a[i]] = dpa[a[i] - 1] + 1;
    }
    dpb[b[1]] = 1;
    for (int i = 2; i <= m; ++i) {
        dpb[b[i]] = dpb[b[i] - 1] + 1;
    }
    int ans = -inf;
    for (int i = 1; i <= mx; ++i) {
        ans = max(ans, min(dpa[i], dpb[i]));
    }
    printf("%d\n", ans);
    return ;
}
int main() {
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}