The 2018 ACM-ICPC Asia Beijing Regional Contest

Posted 2021-02-15 ruthank

训练时间：2019-03-31

本场阿渠连出A和D，成功带我们晋级。

I题我坚定的写Java，完全没往打表找规律上想。背锅。

 

A - Jin Yong’s Wukong Ranking List (HihoCoder - 1870)

给你n对拓扑关系，找出第一个不符合之前的拓扑关系的拓扑对。

建图，每加入一对拓扑对 x -> y，看看从 y 是否能跑到 x 即可。

#include <bits/stdc++.h>
#define FOPI freopen("in.txt", "r", stdin);
#define FOPO freopen("out.txt", "w", stdout);
using namespace std;
typedef long long LL;
const int maxn = 2000 + 100;

map<string, int> M;
string s[maxn], t[maxn];
vector<int> v[maxn];

void build(int x, int y)
{
    v[x].push_back(y);
}

int n;

bool dfs(int x, int y)
{
    for (int i = 0; i < v[x].size(); i++)
    {
        if (v[x][i] == y) return false;
        if (!dfs(v[x][i], y)) return false;
    }
    return true;
}

bool check()
{
    M.clear();
    for (int i = 1; i <= n*2; i++) v[i].clear();
    int cnt = 0;
    for (int i = 1; i <= n; i++)
    {
        if (!M.count(s[i])) M[s[i]] = ++cnt;
        if (!M.count(t[i])) M[t[i]] = ++cnt;
        int x = M[s[i]], y = M[t[i]];
        build(x, y);
        if (!dfs(y, x))
        {
            cout << s[i] << " " << t[i] << endl;
            return false;
        }
    }
    return true;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    while(cin >> n)
    {
        for (int i = 1; i <= n; i++) cin >> s[i] >> t[i];
        if (check()) cout << 0 << endl;
    }
}

 

B - Heshen‘s Account Book (HihoCoder - 1871)

坑巨多的模拟。幸好场上没开。

 

#include <bits/stdc++.h>
#define FOPI freopen("in.txt", "r", stdin);
#define FOPO freopen("out.txt", "w", stdout);
using namespace std;
typedef long long LL;
const int maxn = 200 + 100;

string str[maxn];
int l[maxn], tot[maxn];
bool vis[maxn];
vector<LL> ans;

bool check(string s)
{
    int len = s.length();
    for (int i = 0; i < len; i++)
        if (!isdigit(s[i])) return false;
    return true;
}


bool check_first(string s)
{
    int len = s.length();
    int flag = 0;
    for (int i = len-1; i >= 0; i--)
    {
        if (isdigit(s[i])) flag++;
        else if (s[i] == ‘ ‘) return flag > 0;
        else return false;
    }
    return true;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    //FOPI;

    memset(vis, false, sizeof(vis));

    int cnt = 0;
    while(getline(cin, str[++cnt]));
    cnt--;

    bool flag = false;
    int i;
    for (i = 1; i <= cnt; i++)
    {
        string s = str[i];
        string last = "";
        int len = s.length();

        int tmp = i;
        if (check_first(s))
        {
            int j;
            for (j = i+1; j <= cnt; j++)
                if (check(str[j])) s += str[j], vis[j] = true, tmp = cnt;
                else { tmp = j-1; break; }
        }

        len = s.length();
        if (i != cnt && isdigit(s[len-1]) && isdigit(str[i+1][0]) && !vis[i+1])
        {
            vis[i+1] = true;
            for (int j = 0; j < str[i+1].length() && str[i+1][j] != ‘ ‘; j++) s += str[i+1][j];
        }

        stringstream ss;
        ss.str(s);

        string a;
        if (vis[i]) ss >> a;
        while(ss >> a)
        {
            int len = a.length();
            if (len > 1 && a[0] == ‘0‘) continue;

            if (isalpha(a[0]) || isalpha(a[len-1])) continue;

            LL sum = 0;
            for (int j = 0; j < len; j++)
                if (isdigit(a[j])) sum = sum * 10 + a[j] - ‘0‘;

            ans.push_back(sum), tot[i]++;
        }

        i = tmp;
        if (i == cnt) break;
    }

    int sz = ans.size();
    for (int i = 0; i < sz-1; i++) printf("%lld ", ans[i]);
    printf("%lld
", ans[sz-1]);

    for (int i = 1; i <= cnt; i++) printf("%d
", tot[i]);
}

 

 

I - Palindromes (HihoCoder - 1878)

打个表，找到规律。

然后分类讨论就行了。

#include <bits/stdc++.h>
#define FOPI freopen("in.txt", "r", stdin);
#define FOPO freopen("out.txt", "w", stdout);
using namespace std;
const int maxn = 100000 + 1000;

int t;
char s[maxn], r[maxn];

int main()
{
    scanf("%d", &t);
    for (int ca = 1; ca <= t; ca++)
    {
        scanf("%s", s);
        int len = strlen(s);

        if (len == 1) { printf("%c
", s[0]-1); continue; }
        if (len == 2 && s[1] == ‘0‘ && s[0] == ‘1‘) { printf("%d
", 9); continue; }

        if (s[0] == ‘1‘)
        {
            if (s[1] == ‘0‘)
            {
                s[1] = ‘9‘;
                for (int i = 1; i < len; i++) printf("%c", s[i]);
                for (int i = len-2; i >= 1; i--) printf("%c", s[i]);
            }
            else
            {
                for (int i = 1; i < len; i++) printf("%c", s[i]);
                for (int i = len-1; i >= 1; i--) printf("%c", s[i]);
            }
        }
        else
        {
            s[0]--;
            printf("%s", s);
            for (int i = len-2; i >= 0; i--) printf("%c", s[i]);
        }

        puts("");
    }
}