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hzau 1205 Sequence Number（二分）
G. Sequence Number
In Linear algebra, we have learned the definition of inversion number: Assuming A is a ordered set with n numbers ( n > 1 ) which are different from each other. If exist positive integers i , j, ( 1 ≤ i ＜ j ≤ n and A[i] ＞ A[j]), <a[i], a[j]=""> is regarded as one of A’s inversions. The number of inversions is regarded as inversion number. Such as, inversions of array <2,3,8,6,1> are <2,1>, <3,1>, <8,1>, <8,6>, <6,1>,and the inversion number is 5. Similarly, we define a new notion —— sequence number, If exist positive integers i, j, ( 1 ≤ i ≤ j ≤ n and A[i] <= A[j], <a[i], a[j]=""> is regarded as one of A’s sequence pair. The number of sequence pairs is regarded as sequence number. Define j – i as the length of the sequence pair. Now, we wonder that the largest length S of all sequence pairs for a given array A.
Input
There are multiply test cases. In each case, the first line is a number N(1<=N<=50000 ), indicates the size of the array, the 2th ~n+1th line are one number per line, indicates the element Ai （1<=Ai<=10^9） of the array.
Output
Output the answer S in one line for each case.
Sample Input
5 2 3 8 6 1
Sample Output
3
题意：找出最远的i<=j&&a[i]<=a[j]的长度；
思路：这是一道排序可以过的题，也可以rmq+二分 最快的写法可以用单调栈做到O(n)
我是求后面的最大值后缀，二分后缀；
1 #pragma comment(linker, "/STACK:1024000000,1024000000") 2 #include<iostream> 3 #include<cstdio> 4 #include<cmath> 5 #include<string> 6 #include<queue> 7 #include<algorithm> 8 #include<stack> 9 #include<cstring>10 #include<vector>11 #include<list>12 #include<set>13 #include<map>14 using namespace std;15 #define ll long long16 #define pi (4*atan(1.0))17 #define eps 1e-418 #define bug(x) cout<<"bug"<<x<<endl;19 const int N=1e5+10,M=1e6+10,inf=2147483647;20 const ll INF=1e18+10,mod=2147493647;21 22 int a[N],nex[N];23 int main()24 {25 int n;26 while(~scanf("%d",&n))27 {28 memset(nex,0,sizeof(nex));29 for(int i=1;i<=n;i++)30 scanf("%d",&a[i]);31 for(int j=n;j>=1;j--)32 nex[j]=max(a[j],nex[j+1]);33 int ans=0;34 for(int i=1;i<=n;i++)35 {36 int s=i,e=n,pos=-1;37 while(s<=e)38 {39 int mid=(s+e)>>1;40 if(nex[mid]>=a[i])41 pos=mid,s=mid+1;42 else e=mid-1;43 }44 ans=max(ans,pos-i);45 }46 printf("%d\n",ans);47 }48 return 0;49 }
单调栈做法，如果当前元素小于栈顶元素入栈，否则依次和栈中元素计算相对距离且取最长的那一个。
1 #include <cstdio> 2 #include <algorithm> 3 using namespace std; 4 struct jj 5 { 6 int pos,x; 7 }a[50000]; 8 int main() 9 {10 int n,i,i1,x,top,maxnum;11 while(scanf("%d",&n)!=EOF)12 {13 top=0;14 maxnum=0;15 for(i=0;n>i;i++)16 {17 scanf("%d",&x);18 if(top==0||a[top-1].x>x)19 {20 a[top].x=x;21 a[top].pos=i;22 top++;23 }24 else25 {26 for(i1=top-1;i1>=0&&a[i1].x<=x;i1--)27 {28 maxnum=max(maxnum,i-a[i1].pos);29 }30 }31 }32 printf("%d\n",maxnum);33 }34 return 0;35 }
双指针做法，维护左指针和右指针，使左指针的值小于等于右指针的值
1 #include <cstdio> 2 #include <vector> 3 #include <cstring> 4 #include <string> 5 #include <cstdlib> 6 #include <iostream> 7 #include <map> 8 #include <cmath> 9 #include <algorithm>10 using namespace std;11 typedef long long LL;12 typedef pair<int,int>pii;13 const int N = 1e5+5;14 const double eps = 1e-8;15 int T,n,w[N],sum[N<<2],p[N<<2],cnt,m,ret[N];16 int k,a[N],mi[N];17 int main() {18 while(~scanf("%d",&n)){19 20 int ans=0;21 mi[0]=1e9+10;22 for(int i=1;i<=n;i++){23 scanf("%d",&a[i]);24 mi[i]=1e9+10;25 }26 for(int i=1;i<=n;i++){27 mi[i]=min(mi[i-1],a[i]);28 }29 for(int l=n,r=n;l>=1;l--){30 if(mi[l]>a[r]){31 while(a[r]<mi[l]){32 r--;33 }34 ans=max(ans,r-l);35 }36 else {37 ans=max(ans,r-l);38 }39 }40 printf("%d\n",ans);41 }42 return 0;43 }
hzau 1205 Sequence Number（二分）