【算法学习笔记】89. 序列型动态规划 SJTU OJ 4020 数列游戏

http://acm.sjtu.edu.cn/OnlineJudge/problem/4020

一上手就来了一个删点 排序+DFS.... 虽然正确性没问题 但是超时 只有60分. 主要在于不知道怎么减少搜索量

思路就是删除一些肯定不能在的点, 然后经过条件判断 DFS地去搜索最长的路径

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <stack>
using namespace std;
int n;
int data[1000+10];
bool vis[1000+10]={false};
struct Point
{
    int oriPos;
    int num;
    int todo;
    int vid;
    int done;
    int len;
    Point(int x,int y){
        oriPos = x;
        num = y;
        todo = oriPos - num;
        vid = -1;
        done = 0;
        len = 0;
    }
};
vector<Point> v;
 
bool cmp_point(const Point& a , const Point& b){
    return a.num < b.num;
}
 
void init(){
    v.clear();
    for (int i = 1; i <= n; ++i)
    {
        cin>>data[i];
        if(data[i]<=i){
            Point t(i,data[i]);
            v.push_back(t);
        }
    }
    sort(v.begin(),v.end(),cmp_point);
    memset(vis,false,sizeof(vis));
}
 
//必须dfs
int build(){
    for (int i = 0; i < v.size(); ++i)
    {
        //cout<<v[i].oriPos<<","<<v[i].num<<","<<v[i].todo<<endl;
        v[i].vid = i;
    }
    
    int ans = 0;
    stack<Point> s;
    Point start(0,0);
    s.push(start);
    while(!s.empty()){
        Point cur = s.top();
        s.pop();
        if(cur.vid >= 0)
            vis[cur.vid] = true;
        for (int j = cur.vid+1; j < v.size(); ++j) {
            if(cur.vid==-1 or (v[j].oriPos > v[cur.vid].oriPos and v[j].todo-cur.done >= 0 and v[j].num!=v[cur.vid].num)){
                vis[j] = true;
                Point next(0,0);
                next.oriPos = v[j].oriPos; next.num = v[j].num;
                next.len = cur.len+1;
                next.done = v[j].todo;
                next.vid = j;
                ans = max(next.len,ans);
                s.push(next);
            }
        }
    }
    return ans;
}
 
int main(int argc, char const *argv[])
{
    while(cin>>n){
        init();
        cout<<build()<<endl;
    }
    return 0;
}

DFS