最长递增子序列

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1. 最长递增子序列

求长度

#include<iostream>#include<algorithm>#include<string>#include<vector>using namespace std;int LIS(int *d, int n) {    vector<int> g(n, 0);    int k = 0;    for(int i = 0; i < n; i++) {        int pos = lower_bound(g.begin(), g.begin()+k, d[i]) - g.begin();        if(pos == k) k++;        g[pos] = d[i];    }    for(int i = 0; i < k; i++) cout<<g[i]<<" ";    cout<<endl;    return k;}int main() {    int n;    while(cin>>n) {        int *d = new int[n];        for(int i = 0; i < n; i++) cin>>d[i];        cout<<LIS(d, n)<<endl;        delete[] d;    }}

求长度与一条最短的序列，这种方法得到的一定是当前字典序最小的方案

#include<iostream>#include<algorithm>#include<string>#include<string.h>#include<vector>using namespace std;int LIS(int *d, int n) {    vector<int> g(n, 0);    vector<int> p(n, 0);    int pre[n];    for(int i = 0; i < n; i++) pre[i] = i;    int k = 0;    for(int i = 0; i < n; i++) {        int pos = lower_bound(g.begin(), g.begin()+k, d[i]) - g.begin();        if(pos == k) k++;        g[pos] = d[i];        p[pos] = i;        if(pos) pre[i] = p[pos-1];    }    int t = p[k-1];    int j = k - 1;    vector<int> ans(k);    while(j >= 0) {        ans[j--] = d[t];        t = pre[t];    }    for(int i = 0; i < k; i++) cout<<ans[i]<<" ";    cout<<endl;    return k;}int main() {    int n;    while(cin>>n) {        int *d = new int[n];        for(int i = 0; i < n; i++) cin>>d[i];        cout<<LIS(d, n)<<endl;        delete[] d;    }}

2. 最长非递减子序列（包含重复元素）

将上述函数的

lower_bound 修改为 upper_bound就可以了

这两个函数包含在头文件#include <algorithm>

3. 最长递增子串（求长度以及输出对应路径）

#include<iostream>#include<algorithm>#include<string>#include<string.h>#include<vector>using namespace std;int LIS(int *d, int n) {    int max_len = 1;    int max_i = 0;    int cur_len = 1;    for(int i = 1; i < n; i++) {        if(d[i] > d[i-1]) cur_len++;        else cur_len = 1;        if(cur_len > max_len) {            max_len = cur_len;            max_i = i;        }    }    for(int i = max_i + 1 - max_len; i <= max_i; i++) cout<<d[i]<<" ";    cout<<endl;    cout<<"max_len = "<<max_len<<endl;    return max_len;}int main() {    int n;    while(cin>>n) {        int *d = new int[n];        for(int i = 0; i < n; i++) cin>>d[i];        cout<<LIS(d, n)<<endl;        delete[] d;    }}

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