最长上升子序列模型 AcWing 1010. 拦截导弹

叁歲伎倆 2023-09-28 14:04 8阅读 0赞

## 最长上升子序列模型 AcWing 1010. 拦截导弹 ##

### 原题链接 ###

[AcWing 1010. 拦截导弹][AcWing 1010.]

### 算法标签 ###

DP 线性DP 最长上升子序列

### 思路 ###

![在这里插入图片描述][55cabe352b6c404fb267f325768508e1.png]  
[摘自该题解][Link 1]

### 代码 ###

#include<bits/stdc++.h>
    #define int long long
    #define rep(i, a, b) for(int i=a;i<b;++i)
    #define Rep(i, a, b) for(int i=a;i>=b;--i)
    using namespace std;
    const int N = 1005, INF = 0x3f3f3f3f;
    // f1[i]表示此时第i套导弹拦截系统所拦截的最后一个导弹的高度
    int f[N], f1[N], a[N];
    inline int read(){
       int s=0,w=1;
       char ch=getchar();
       while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
       while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
       return s*w;
    }
    void put(int x) {
        if(x<0) putchar('-'),x=-x;
        if(x>=10) put(x/10);
        putchar(x%10^48);
    }
    vector<string> get(string str){
        vector<string> res;
        string word;
        for (auto c: str){
            if (c == ' '){
                res.push_back(word);
                word = "";
            }
            else word += c;
        }
        res.push_back(word);
        return res;
    }
    signed main(){
    	ios::sync_with_stdio(false);
    	cin.tie(0);
    	cout.tie(0);
    	string s;
    	getline(cin, s);
    	vector<string> v=get(s);
    	int ans=0, cnt=0, n=v.size();// cnt代表现有子序列长度
    	rep(i, 0, n){
    	    a[i]=stoi(v[i]);
    	}
    	// 以i结尾下降序列
    	rep(i, 0, n){
    	    f[i]=1;
    	    rep(j, 0, i){
    	        if(a[j]>=a[i]){
    	            f[i]=max(f[i], f[j]+1);
    	        }
    	    }
    	    ans=max(ans, f[i]);
    	    int k=0;
            while(k<cnt&&f1[k]<a[i]){
                k++;    
            } 
            // a[i]开创一套新拦截系统   
            if(k==cnt){
                f1[cnt++]=a[i];  
            } 
            // a[i]成为第k套拦截系统最后一个导弹高度
            else{
                f1[k]=a[i];    
            }
    	}
    	printf("%lld\n%lld\n", ans, cnt);
    }

### y总代码 ###

#include <sstream>
    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    const int N = 1010;
    
    int n;
    int h[N], f[N], q[N];
    
    int main()
    {
        string line;
        getline(cin, line);
        stringstream ssin(line);
        while (ssin >> h[n]) n ++ ;
    
        int res = 0, cnt = 0;
        for (int i = 0; i < n; i ++ )
        {
            f[i] = 1;
            for (int j = 0; j < i; j ++ )
                if (h[i] <= h[j])
                    f[i] = max(f[i], f[j] + 1);
            res = max(res, f[i]);
    
            int k = 0;
            while (k < cnt && q[k] < h[i]) k ++ ;
            if (k == cnt) q[cnt ++ ] = h[i];
            else q[k] = h[i];
        }
    
        printf("%d\n", res);
        printf("%d\n", cnt);
        return 0;
    }

### tips ###

读入处理  
输入流高级语法

string line;
        getline(cin, line);
        stringstream ssin(line);
        while (ssin >> h[n]) n ++ ;

或

while (cin >> h[ ++ n]);

[AcWing 1010.]: https://www.acwing.com/problem/content/1012/
[55cabe352b6c404fb267f325768508e1.png]: https://img-blog.csdnimg.cn/55cabe352b6c404fb267f325768508e1.png
[Link 1]: https://www.acwing.com/solution/content/10173/