LIS(n^2) POJ 2533 Longest Ordered Subsequence
题意:LIS(Longest Increasing Subsequence)裸题
分析:状态转移方程:dp[i] = max (dp[j]) + 1 (a[j] < a[i],1 <= j < i) 附带有print输出路径函数
代码:
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std; const int N = 1e4 + 10;
const int INF = 0x3f3f3f3f;
int a[N], dp[N], fa[N];
int n; void print(int x) {
if (fa[x] != -1) {
print (fa[x]); printf (" %d", a[x]);
}
else printf ("%d", a[x]);
} void LIS(void) {
int ret = 0, last = 0;
for (int i=1; i<=n; ++i) {
dp[i] = 1; fa[i] = -1;
for (int j=1; j<i; ++j) {
if (a[j] < a[i] && dp[i] < dp[j] + 1) {
dp[i] = dp[j] + 1; fa[i] = j;
}
}
if (ret < dp[i]) {
ret = dp[i]; last = i;
}
} printf ("%d\n", ret);
// print (last); puts ("");
} int main(void) {
while (scanf ("%d", &n) == 1) {
for (int i=1; i<=n; ++i) {
scanf ("%d", &a[i]);
}
LIS ();
} return 0;
}
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