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 ", ans); return ; } int main() { #ifdef local freopen("data.txt","r",stdin); #endif int t; scanf("%d", &t); while (t--) work(); return 0; }