NOI 2387:Til the Cows Come Home

题目

http://bailian.openjudge.cn/practice/2387/

思路

狄杰斯特拉,优先级队列和 d 数组保持同步即可,使用邻接表的话,Edge 中的 from 字段可以省略

源码

//
//  2387.cpp
//  test
//
//  Created by bytedance on 2020/9/8.
//  Copyright © 2020 bytedance. All rights reserved.
//

#include <stdio.h>
#include <iostream>
#include <vector>
#include <queue>
#define inf 99999999
using namespace std;
struct Edge{
    int from,to,dis;
    Edge(int from,int to,int dis):from(from),to(to),dis(dis){};
    bool operator < (const Edge &e) const{
        return e.dis<dis;
    }
};

int n, t;

int main(){
    cin >> t >> n;
    vector<vector<Edge>> edges(n+1,vector<Edge>());
    priority_queue<Edge> q;
    vector<int> d(n+1,inf);
    vector<bool> done(n+1);
    while(t--){
        int a,b,c;
        cin >> a>>b>>c;
        Edge e(a,b,c);
        edges[a].push_back(e);
        Edge e2(b,a,c);
        edges[b].push_back(e2);
    }
    //边读入ok,初始化距离数组
    q.push(Edge(1,1,0));
    d[1]=0;
    while(!q.empty()){
        int node  = q.top().to;q.pop();
        if(done[node])continue;
        done[node] = true;
        for(int i = 0;i < edges[node].size();++i){
            Edge e = edges[node][i];
            if(d[e.to]>d[node]+e.dis){
                d[e.to] = d[node]+e.dis;
                q.push(Edge(1,e.to,d[e.to]));
            }
        }
    }
    cout << d[n];
    return 0;
}

NOI 1840:Eqs

题目

http://bailian.openjudge.cn/practice/1840/

思路

把等式右移三项,遍历左边的两项可能的取值用 map 存下来,遍历右边值而后从map中取出对应个数相加

源码

//
//  1840.cpp
//  test
//
//  Created by bytedance on 2020/9/8.
//  Copyright © 2020 bytedance. All rights reserved.
//

#include <stdio.h>
#include <iostream>
#include <map>
#include <math.h>
using namespace std;

int main(){
    int cnt = 0;
    map<int,int> m;
    int a1,a2,a3,a4,a5;
    cin >> a1>>a2>>a3>>a4>>a5;
    for(int i = -50;i <= 50;++i){
        int res1 = i*i*i*a1;
        for(int j = -50;j <= 50;++j){
            int res = res1+j*j*j*a2;
            if(i==0||j==0)continue;
            if(m.find(res)!=m.end()){
                m[res]++;
            }else{
                m[res] = 1;
            }

        }
    }
    //    for(map<int,int>::iterator it = m.begin();it!=m.end();++it){
    //        cout <<it->first<<" "<<it->second<<endl;
    //    }

    for(int i = -50;i <= 50;++i){
        for(int j = -50;j <=50;++j){
            for(int k = -50;k <= 50;++k){
                if(i==0||j==0||k==0)continue;
                int res = i*i*i*a3+j*j*j*a4+k*k*k*a5;
                res = -res;
                if(m.find(res)!=m.end()){
                    cnt+=m[res];
                }
            }
        }
    }
    cout << cnt;
    return 0;
}

POJ 3126.Prime Path 素数筛+双向BFS

题目

http://poj.org/problem?id=3126

思路

素数筛给出4位素数,双向 BFS

源码

//
//  3126.cpp
//  test
//
//  Created by bytedance on 2020/9/8.
//  Copyright © 2020 bytedance. All rights reserved.
//
#include <stdio.h>
#include <iostream>
#include <vector>
#include <queue>
#include <map>
using namespace std;
bool pre[10000]={false};
vector<string> p;
int main(){
    int time;
    cin >> time;

    string s,e;
    for(int i = 2;i <= 9999;++i){
        for(int j = i*2;j<=9999;j+=i)pre[j]=true;
    }
    for(int i = 1000;i<=9999;++i){
        if(!pre[i])p.push_back(to_string(i));
    }
    while(time--){
        cin >> s>>e;
        //素数集合构造完毕
        int idx = 0;
        queue<string> q[2];
        map<string,bool> vis[2];
        q[0].push(s);
        q[1].push(e);
        vis[0][s]=true;
        vis[1][e]=true;
        int cnt = 0;
        bool f = true;
        while(!q[0].empty()&&!q[1].empty()&&f){
            idx=idx^1;
            int sz = q[idx].size();
            while(sz--&&f){
                string crt = q[idx].front();
                q[idx].pop();
                if(vis[idx^1][crt]==true){ cout<< cnt; f=false;}
                for(int i = 0;i < 4&&f;++i){
                    char c = crt[i];
                    for(int j = '0';j <='9'&&f;++j){
                        crt[i] = j;
                        if(j==c||vis[idx][crt]||find(p.begin(),p.end(),crt)==p.end())continue;
                        q[idx].push(crt);
                        vis[idx][crt]=true;
                    }
                    crt[i] = c;
                }
            }
            cnt++;
        }
    }
    return 0;
}

NOI 2355:Railway tickets dp

题目

http://bailian.openjudge.cn/practice/2355/

思路

我觉得可以用 狄杰斯特拉但是我写完 wa 了不知道为啥
dp[i]表示到i的最短路径,直接 dp 求解,i 从 1到n遍历求解dp[i],只会用到 dp[k] k < i

源码

//
//  2355.cpp
//  test
//
//  Created by bytedance on 2020/9/7.
//  Copyright © 2020 bytedance. All rights reserved.
//

#include <iostream>
#include <vector>
#define INF 999999999999
using namespace std;

long c1,c2,c3,l1,l2,l3;
int n,s,e;
long pre[10005]={0};
int main(){
    cin >>l1>>l2>>l3;
    cin >> c1>>c2>>c3;
    cin >> n;
    cin >>s>>e;
    if(s>e)swap(s,e);
    for(int i = 2;i <= n;++i){
        cin >> pre[i];
    }
    vector<long>d(n+1,INF);
    d[s] = 0;
    for(int i = s+1;i <= n;++i){
        for(int j = i-1;j>=s;--j){
            long dis = pre[i]-pre[j];
            if(dis<=l1)d[i]=min(d[i],d[j]+c1);
            else if(dis<=l2)d[i] = min(d[i],d[j]+c2);
            else if(dis<=l3)d[i] = min(d[i],d[j]+c3);
            else break;
        }
    }
    cout<<d[e];
    return 0;
}