POJ 2533 - Longest Ordered Subsequence
2017-09-06 15:34
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题目大意:求最长递增子序列的长度。
解题思路:每个dp初始为1。第一个循环位置从0到n,判断,如果之后的有比该位置大的,就去判断后位置的dp大小。dp[j] = max(dp[j], dp[i]+1);
ac代码:
解题思路:每个dp初始为1。第一个循环位置从0到n,判断,如果之后的有比该位置大的,就去判断后位置的dp大小。dp[j] = max(dp[j], dp[i]+1);
ac代码:
#include <iostream> #include <algorithm> using namespace std; int dp[1005], n, a[1005], Max; int main() { while (scanf("%d", &n) != EOF){ for (int i=0; i<n; i++){ scanf("%d", &a[i]); dp[i] = 1 4000 ; } for (int i=0; i<n; i++) for (int j=i+1; j<n; j++) if (a[i] < a[j]) dp[j] = max(dp[j], dp[i]+1); Max = dp[0]; for (int i=1; i<n; i++) Max = max(Max, dp[i]); printf("%d\n", Max); } return 0; }
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