http://acm.hdu.edu.cn/showproblem.php?pid=4604
想了半天，发现是LIS，O（nlgn）飘过。
具体思路：预处理好以i为开头的（下面我用逆序输入，所以是以i为结尾的）LIS和LDS。
再枚举每个点找最优即可（注意不严格递增、减）。
1 #include<map> 2 #include<set> 3 #include<list> 4 #include<cmath> 5 #include<ctime> 6 #include<queue> 7 #include<stack> 8 #include<cctype> 9 #include<cstdio> 10 #include<string> 11 #include<vector> 12 #include<cstdlib> 13 #include<cstring> 14 #include<iostream> 15 #include<algorithm> 16 #define MAXN 100005 17 using namespace std; 18 int num[MAXN], n; 19 int dpI[MAXN], lenI[MAXN]; //I:increasing 20 int dpD[MAXN], lenD[MAXN]; //D:decreasing 21 int cntI[MAXN], cntD[MAXN]; 22 //cnt[i]记录以i结尾的子序列中num[i]的重复个数。 23 24 int searchI(int x) 25 { 26 //printf("done "); 27 int l=1, r=n, m; 28 while(l <= r) 29 { 30 m = (l+r)/2; 31 if(x >= lenI[m]) 32 l = m+1; 33 else 34 r= m-1; 35 } 36 return l-1; 37 } 38 39 int searchD(int x) 40 { 41 int l=1, r=n, m; 42 while(l <= r) 43 { 44 m = (l+r)/2; 45 if(x <= lenD[m]) 46 l = m+1; 47 else 48 r= m-1; 49 } 50 return l-1; 51 } 52 53 void print(int *x) 54 { 55 for(int i=1; i<=n; i++) 56 printf("%d ", x[i]); 57 printf(" "); 58 } 59 60 int main() 61 { 62 //hdu 1005 63 int t; 64 cin >> t; 65 while(t--){ 66 cin >> n; 67 for(int i=n; i>=1; i--) 68 scanf("%d", &num[i]); 69 //为了方便处理，逆序输入。 70 memset(lenI, 0x3f, sizeof(lenI)); 71 for(int i=1; i<=n; i++){ 72 lenD[i] = -0x3f3f3f3f; 73 cntI[i] = cntD[i] = 1; 74 } 75 76 for(int i=1; i<=n; i++){ 77 int k = searchI(num[i]); 78 dpI[i] = k + 1; 79 if(lenI[k] == num[i]) 80 cntI[k+1] = cntI[k]+1; 81 lenI[dpI[i]] = min(lenI[dpI[i]], num[i]); 82 } 83 84 for(int i=1; i<=n; i++){ 85 int k = searchD(num[i]); 86 dpD[i] = k + 1; 87 if(lenD[k] == num[i]) 88 cntD[k+1] = cntD[k]+1; 89 lenD[dpD[i]] = max(lenD[dpD[i]], num[i]); 90 } 91 //print(dpI); 92 //print(dpD); 93 94 int ans = 0; 95 for(int i=1; i<=n; i++) 96 ans = max(ans, dpI[i]+dpD[i]-min(cntI[i], cntD[i])); 97 printf("%d ", ans); 98 99 } 100 101 return 0; 102 }
可以做一下类似题：EOJ 1237 O（n^2）水过