UVa 10534 Wavio Sequence (黑书例题,DP)
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Problem links: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1475
最长上升子序列问题。由于N<=10000,于是要用O(nlogn)的做法。题目要求求出一段子序列,长度为2*k-1,满足先严格上升,再严格下降。所以做法是顺着求一次LIS,反着再求一次。但是O(nlogn)的做法中,dp[i]表示的是最长上升子序列长度为i时,结尾的数是哪个。考虑到对于原序列中任何一个位置x,如果知道从左到右到达它的最长上升子序列,以及从右到左到它的最长上升子序列,枚举每个x,自然可以得到答案。所以在O(nlogn)求序列的时候,顺道记录每个点的左右两端可达长度即可。
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define SIZE 10010using namespace std;int n;int arr[SIZE];int dp[SIZE];int p1[SIZE],p2[SIZE];int binarySearch(int l,int r,int tar){ while (l <= r) { int mid = (l + r) >> 1; if(dp[mid] >= tar) r = mid - 1; else l = mid + 1; } return l;}int main(){ while(~scanf("%d",&n)) { for(int i=1; i<=n; i++) scanf("%d",&arr[i]); int len = 0; dp[++len] = arr[1], p1[1] = 1; for(int i=2; i<=n; i++) { if(arr[i] > dp[len]) dp[++len] = arr[i], p1[i] = len; else { int pos = binarySearch(1,len,arr[i]); dp[pos] = arr[i]; p1[i] = pos; } } len = 0; dp[++len] = arr[n], p2[n] = 1; for(int i=n-1; i>=1; i--) { if(arr[i] > dp[len]) dp[++len] = arr[i], p2[i] = len; else { int pos = binarySearch(1,len,arr[i]); dp[pos] = arr[i]; p2[i] = pos; } } int ans = 0; for(int i=1; i<=n; i++) ans = max(ans,2*min(p1[i],p2[i])); printf("%d\n",ans-1); } return 0;}
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