PTA 1139 1138 1137 1136

PAT 1139 1138 1137 1136

一个月不写题，有点生疏。。脑子跟不上手速，还可以啦，反正今天很开心。

PAT 1139 First Contact

18/30 找个时间再修bug
图论

#include<bits/stdc++.h>
using namespace std;

const int maxn = 9999;
vector<int> g[maxn];
set<int> se[maxn];
int gender[10010];
int n,m;
vector<pair<int,int> >ans;

void init(){
    for(int i=1;i<=10000;i++) gender[i] = 0;
}

bool cmp(pair<int,int> a,pair<int,int> b){
    if(a.first == b.first) return a.second < b.second;
    return a.first < b.first;
}

int main(){
    cin>>n>>m;
    init();
    for(int i=1;i<=m;i++){
        int u,v;
        cin>>u>>v;
        int absu = abs(u);
        int absv = abs(v);
        if(u < 0) gender[absu] = 1;
        if(v < 0) gender[absv] = 1;
        g[absu].push_back(absv);
        g[absv].push_back(absu);
        se[absu].insert(absv);
        se[absv].insert(absu);
    }
    int k;
    cin>>k;
    for(int i=1;i<=k;i++){
        int u,v;
        cin>>u>>v;
        int absu = abs(u);
        int absv = abs(v);
        int genderu = gender[absu];
        int genderv = gender[absv];
        int cnt = 0;
        for(int c=0;c<g[absu].size();c++){
            if(gender[g[absu][c]] != genderu ) continue;
            for(int d=0;d<g[absv].size();d++){
                if(gender[g[absv][d]] != genderv ) continue;
                if(se[g[absv][d]].find(g[absu][c]) == se[g[absv][d]].end()) continue;
                if(se[g[absu][c]].find(g[absv][d]) == se[g[absu][c]].end()) continue;
                cnt++;
                ans.push_back(make_pair(g[absu][c],g[absv][d]));
            }
        }
        cout<<cnt<<endl;
        sort(ans.begin(),ans.end(),cmp);
        for(int i=0;i<cnt;i++){
            cout<<ans[i].first<<" "<<ans[i].second<<endl;
        }
        ans.clear();
    }
    return 0;
}

1138 Postorder Traversal

数据结构，二叉树先序中序建树后后序遍历

#include<bits/stdc++.h>
using namespace std;

const int maxn = 50010;
int n;
int pre[maxn];
int in[maxn];
int post[maxn];
bool flag = false;

struct node{
    int v;
    struct node *l;
    struct node *r;
};

//建树
node* build(int l,int r,int rootIdx){
    if(l > r) return NULL;
    node *root = new node();
    root->v = pre[rootIdx];
    int pos = l;
    while(in[pos] != pre[rootIdx]) pos++;
    root->l = build(l,pos-1,rootIdx+1);
    root->r = build(pos+1,r,rootIdx+(pos-l+1));
    return root;
}

//后序遍历 打印第一个被遍历的 结点值
void postTravel(node *Root){
    if(flag) return;
    if(Root->l == NULL && Root->r == NULL){
        cout<<Root->v<<endl;
        flag = true;
        return;
    }
    if(Root->l) postTravel(Root->l);
    if(Root->r) postTravel(Root->r);
    return;
}

int main(){
    cin>>n;
    for(int i=1;i<=n;i++) cin>>pre[i];
    for(int i=1;i<=n;i++) cin>>in[i];
    node *Root = new node();
    Root = build(1,n,1); //建树
    postTravel(Root);
    return 0;
}

1137 Final Grading

数据结构，map，排序常见题型

#include<bits/stdc++.h>
#include<unordered_map>
using namespace std;

const int maxn = 10010;
int p; //P the number of students having done the online programming assignmen
int m; //the number of students on the mid-term list
int n; //the number of students on the final exam list

struct student{
    string id;
    int gp;
    int gmid;
    int gfinal;
    int g;
};

unordered_map<string,student > mp;
vector<student> ans;

bool cmp(student stua,student stub){
    if(stua.g == stub.g){
        return stua.id < stub.id;
    }
    return stua.g > stub.g;
}

void print(){
    for(int i=0;i<ans.size();i++){
        cout<<ans[i].id<<" "<<ans[i].gp<<" "<<ans[i].gmid<<" "<<ans[i].gfinal<<" "<<ans[i].g<<endl;
    }
}

void input(){
    cin>>p>>m>>n;
    for(int i=1;i<=p;i++){
        string id;
        int gp;
        cin>>id>>gp;
        mp[id].id = id;
        mp[id].gp = gp;
        mp[id].gmid = -1;
        mp[id].gfinal = -1;
        mp[id].g = -1;
    }
    for(int i=1;i<=m;i++){
        string id;
        int gmid;
        cin>>id>>gmid;
        mp[id].gmid = gmid;
    }
    for(int i=1;i<=n;i++){
        string id;
        int gfinal;
        cin>>id>>gfinal;
        mp[id].gfinal = gfinal;
    }
}

int main(){
    input(); 
    for(auto it = mp.begin();it!=mp.end();it++){
        if(it->second.gp < 200) continue;
//      if(it->second.gfinal < 60 || it->second.gfinal > 100) continue; //吼吼 一开始题意理解错啦 g才是final score 
        if(it->second.gmid > it->second.gfinal) {
            int tempScore = ((it->second.gmid*4 + it->second.gfinal*6) );
            if(tempScore % 10 >= 5) it->second.g = tempScore/10+1;
            else it->second.g= tempScore/10;    
        }
        else it->second.g = it->second.gfinal;
        if(it->second.g >= 60 && it->second.g <=100) ans.push_back(it->second);
    }
    sort(ans.begin(),ans.end(),cmp);
    print();
    return 0;
} 

1136 A Delayed Palindrome

大数模拟加法进位

#include<bits/stdc++.h>
using namespace std;

vector<int> v1;
vector<int> v2;
vector<int> ans;
/*
    reverse
    add
*/

void print(){
    for(int i=0;i<v1.size();i++) cout<<v1[i];
    cout<<" + ";
    for(int i=0;i<v2.size();i++) cout<<v2[i];
    cout<<" = ";
    for(int i=0;i<ans.size();i++) cout<<ans[i];
    cout<<endl; 
} 

bool check(){
    int len = ans.size();
    for(int i=0;i<len/2;i++){
        if(ans[i] != ans[len-i-1]) return false;
    }
    return true;
}

void clear(){
    v1.clear();
    v2.clear();
    for(int i=0;i<ans.size();i++) v1.push_back(ans[i]);
    ans.clear();
}
    
int main(){
    int d;
    while(scanf("%1d",&d)!=EOF) v1.push_back(d);
    int t = 0;
    bool flag = false;
    while((t++) < 10){
        for(int i=0;i<v1.size();i++) ans.push_back(v1[i]);
        if(check()){
            for(int i=0;i<ans.size();i++) cout<<ans[i];
            cout<<" is a palindromic number.";
            flag = true;
            break;
        }
        ans.clear();
        int ci = 0;
        for(int i=0;i<v1.size();i++) v2.push_back(v1[i]);
        reverse(v2.begin(),v2.end());
        //大数加法模拟
        for(int i=v1.size()-1;i>=0;i--){
            int si = v1[i] + v2[i] + ci;
            if(si >= 10) {
                ci = si/10; 
                si = si%10;
            }else{
                ci = 0;
            }
            ans.push_back(si);
        }
        if(ci!=0) ans.push_back(ci);
        reverse(ans.begin(),ans.end());//反转才是 结果和
        ci = 0;
        print();
        if(check()){
            for(int i=0;i<ans.size();i++) cout<<ans[i];
            cout<<" is a palindromic number.";
            flag = true;
            break;
        }
        clear();
    }
    if(flag == false) cout<<"Not found in 10 iterations."<<endl;
    return 0;
}