POJ 2127 Greatest Common Increasing Subsequence（DP，LCIS）

题意：求最长上升公共子序列

思路：和LCS差不多的状态转移，如果a[i] == b[j]，那么就要往前找到一个dp[i - 1][[k]满足b[j] > b[k]的最大值，时间复杂度是N^3，但是其实可以在遍历j层状态的时候，顺便记录下《 b[j]的dp[i - 1][k]最大值即可

代码：

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

const int N = 505;

int n, m, a
, b
, dp

, path

, out
, on;

int bo = 0;

void print(int n, int m) {
if (n == 0)
return;
print(n - 1, path
[m]);
if (path
[m] != m) {
if (bo) printf(" ");
else bo = 1;
printf("%d", b[m]);
}
}

int main() {
while (~scanf("%d", &n)) {
bo = 0;
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]);
memset(dp, 0, sizeof(dp));
for (int i = 1; i <= n; i++) {
int Max = 0, Maxv = 0;
for (int j = 1; j <= m; j++) {
if (dp[i][j] < dp[i - 1][j]) {
dp[i][j] = dp[i - 1][j];
path[i][j] = j;
}
if (a[i] == b[j]) {
if (dp[i][j] < Max + 1) {
dp[i][j] = Max + 1;
path[i][j] = Maxv;
}
}
if (b[j] < a[i]) {
if (Max < dp[i - 1][j]) {
Max = dp[i - 1][j];
Maxv = j;
}
}
}
}
int ansv = 1;
for (int i = 1; i <= m; i++) {
if (dp
[ansv] < dp
[i]) {
ansv = i;
}
}
printf("%d\n", dp
[ansv]);
print(n, ansv);
printf("\n");
}
return 0;
}