You are given an integer array of length nn.
You have to choose some subsequence of this array of maximum length such that this subsequence forms a increasing sequence of consecutive integers. In other words the required sequence should be equal to [x,x+1,…,x+k?1][x,x+1,…,x+k?1] for some value xx and length kk.
Subsequence of an array can be obtained by erasing some (possibly zero) elements from the array. You can erase any elements, not necessarily going successively. The remaining elements preserve their order. For example, for the array [5,3,1,2,4][5,3,1,2,4] the following arrays are subsequences: [3][3], [5,3,1,2,4][5,3,1,2,4], [5,1,4][5,1,4], but the array [1,3][1,3] is not.
The first line of the input containing integer number nn (1≤n≤2?1051≤n≤2?105) — the length of the array. The second line of the input containing nn integer numbers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the array itself.
On the first line print kk — the maximum length of the subsequence of the given array that forms an increasing sequence of consecutive integers.
On the second line print the sequence of the indices of the any maximum length subsequence of the given array that forms an increasing sequence of consecutive integers.
7 3 3 4 7 5 6 8
4 2 3 5 6
6 1 3 5 2 4 6
2 1 4
4 10 9 8 7
1 1
9 6 7 8 3 4 5 9 10 11
6 1 2 3 7 8 9
All valid answers for the first example (as sequences of indices):
All valid answers for the second example:
All valid answers for the third example:
All valid answers for the fourth example:
题意：给你一个数组找出最长的递增子序列的长度以及下标位置。
例如第一组样例：7
3 3 4 7 5 6 8
最长的子序列为3 4 5 6，长度为4。
下标为1 3 5 6或2 3 5 6
题解：用map进行动态规划，mp[i]表示以i数字开头的最长的递增子序列的长度，即有转移方程mp[i]=max(mp[i],mp[i-1]+1)
最后找出map里面子序列长度最长的数字i，倒着输出就行了。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<string> #include<algorithm> #include<vector> #include<queue> #include<set> #include<map> #include<stack> using namespace std; const int inf=0x3f3f3f3f; map<int ,int >dp; int t[300005]; stack<int>s; int main() { int n; scanf("%d",&n); int ans=-inf; int maxx; for(int i=1;i<=n;i++) { scanf("%d",&t[i]); dp[t[i]]=dp[t[i]-1]+1; if(dp[t[i]]>ans) { ans=dp[t[i]]; maxx=t[i]; } } printf("%d\n",ans); for(int i=n;i>=1;i--) { if(t[i]==maxx) { s.push(i); maxx--; } } for(int i=0;i<ans;i++) { if(i!=0)printf(" "); printf("%d",s.top()); s.pop(); } printf("\n"); }
codeforce 977 F. Consecutive Subsequence
原文：https://www.cnblogs.com/caiyishuai/p/9021456.html