HDU-5748 Bellovin 【LIS（STL应用）】

E - Bellovin
Time Limit:3000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I64u
Submit

Status
Description
Peter has a sequence and he define a function on the sequence – , where is the length of the longest increasing subsequence ending with .

Peter would like to find another sequence in such a manner that equals to . Among all the possible sequences consisting of only positive integers, Peter wants the lexicographically smallest one.

The sequence is lexicographically smaller than sequence , if there is such number from to , that for and .
Input
There are multiple test cases. The first line of input contains an integer , indicating the number of test cases. For each test case:

The first contains an integer – the length of the sequence. The second line contains integers .
Output
For each test case, output integers denoting the lexicographically smallest sequence.

Sample Input
3
1
10
5
5 4 3 2 1
3
1 3 5

Sample Output
1
1 1 1 1 1
1 2 3

---

1. 题意：给定序列a1,a2,a2…..aN,f(a i)代表以a i为结尾元素的最大上升子序列长度，让你求出一个序列b1，b2….bN，使任意的f(a i)=f(b i),求出满足要求的最小序列并输出(b i>1，从前向后贪心满足最小的)；
2. 思路：要输出最小的，那第一个肯定是1，第二个就要看f(a2),如果f(a2)=1那么第二个肯定选1，如果不成立，那么为2，以此类推可以知道bi=f(bi)=f(ai），就是求出每一位的最长上升子序列的长度；
3. 失误：用vector写不知道哪错了，指针一直不对，调试好久也找不出来，最后一急把代码调试的一删，没想到就对了，错误还能这样找；
4. 代码如下：

---

#include<cstdio>#include<vector>using namespace std;int main(){    int T,N,i,tem;      scanf("%d",&T);    while(T--)    {        scanf("%d",&N);        scanf("%d",&tem);   vector<int> VEC,ord; vector<int>::iterator iter;        VEC.push_back(tem); ord.push_back(1);          for(i=2;i<=N;++i)        {            scanf("%d",&tem);            iter = lower_bound(VEC.begin(),VEC.end(),tem);             ord.push_back( (iter-VEC.begin())+1);//用队列可能会更方便             if(iter==VEC.end()) VEC.push_back(tem);            else *iter = tem;          }         int k=0;        for(iter=ord.begin();iter!=ord.end();++iter)        {            if(k!=0) printf(" "); k=1;            printf("%d",*iter);        }        printf("\n");    }    return 0;}