ICPC North Central NA Contest 2017

2024-03-17 09:36:52

ICPC North Central NA Contest 2017

C. Urban Design

  • 思路：用直线的一般式判断两直线相交 通过奇偶性判断是否在同一区域

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 1e4 + 10;

int s, t, x1, y1_, x2, y2, x3, y3, x4, y4, cnt;
int a[N], b[N], c[N];
ll tmp1, tmp2;

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> s;
    for (int i = 1; i <= s; i ++ ){
        cin >> x1 >> y1_ >> x2 >> y2;
        a[i] = y1_ - y2;
        b[i] = x2 - x1;
        c[i] = x1 * y2 - y1_ * x2;
    }
    cin >> t;
    while (t -- ){
        cnt = 0;
        cin >> x3 >> y3 >> x4 >> y4;
        for (int i = 1; i <= s; i ++ ){
            tmp1 = 1ll * x3 * a[i] + 1ll * y3 * b[i] + c[i],
            tmp2 = 1ll * x4 * a[i] + 1ll * y4 * b[i] + c[i];
            if (tmp1 * tmp2 < 0)
                cnt ++ ;
        }
        if (cnt & 1)
            cout << "different\n";
        else
            cout << "same\n";
    }
    return 0;
}

E. Is-A? Has-A? Who Knowz-A?

  • 思路：floyd算法（忘了init导致wa1

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 510;

int n, m, tot;
bool Is[N][N], Has[N][N];
string a, b, c;
map<string, int> mp;

void init(){
    for (int i = 1; i <= tot; i ++ )
        Is[i][i] = true;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i ++ ){
        cin >> a >> b >> c;
        if (!mp.count(a))
            mp[a] = ++ tot ;
        if (!mp.count(c))
            mp[c] = ++ tot ;
        if (b == "is-a")
            Is[mp[a]][mp[c]] = true;
        else
            Has[mp[a]][mp[c]] = true;
    }
    init();
    for (int k = 1; k <= tot; k ++ ){
        for (int i = 1; i <= tot; i ++ ){
            for (int j = 1; j <= tot; j ++ ){
                if (Is[i][k] && Is[k][j])
                    Is[i][j] = true;
                if ((Has[i][k] && Has[k][j]) || (Has[i][k] && Is[k][j]) || (Is[i][k] && Has[k][j]))
                    Has[i][j] = true;
            }
        }
    }
    for (int i = 1; i <= m; i ++ ){
        cin >> a >> b >> c;
        if (b == "is-a"){
            if (Is[mp[a]][mp[c]])
                cout << "Query " << i << ": true\n";
            else
                cout << "Query " << i << ": false\n";
        }
        else{
            if (Has[mp[a]][mp[c]])
                cout << "Query " << i << ": true\n";
            else
                cout << "Query " << i << ": false\n";
        }
    }
    return 0;
}

G. Sheba's Amoebas

  • 思路：dfs

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 110;
const int dx[8] = {0, 0, 1, -1, 1, 1, -1, -1};
const int dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};

int n, m, ans;
char mp[N][N];

void dfs(int x, int y){
    mp[x][y] = '.';
    for (int i = 0; i < 8; i ++ ){
        int tx = x + dx[i],
            ty = y + dy[i];
        if (x >= 0 && x < n && y >= 0 && y < m && mp[tx][ty] == '#')
            dfs(tx, ty);
    }
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < m; j ++ )
            cin >> mp[i][j];
    for (int i = 0; i < n; i ++ ){
        for (int j = 0; j < m; j ++ ){
            if (mp[i][j] == '#'){
                dfs(i, j);
                ans ++ ;
            }
        }
    }
    cout << ans << "\n";
    return 0;
}

H. Zebras and Ocelots

  • 思路：水题 直接用pow会wa（迷惑

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 65;

int n;
ll ans;
ll pow_2[N];
char c[N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    pow_2[0] = 1;
    for (int i = 1; i < N; i ++ )
        pow_2[i] = 2 * pow_2[i - 1];
    cin >> n;
    for (int i = n; i >= 1; i -- )
        cin >> c[i];
    for (int i = 1; i <= n; i ++ )
        if (c[i] == 'O')
            ans += pow_2[i - 1];
    cout << ans << "\n";
    return 0;
}

I. Racing Around the Alphabet

  • 思路：水题 注意处理带空格的串

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const double pi = acos(-1.0);
const int N = 130;

int n, tmp1, tmp2;
double res, ans;
char s[N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    cin.getline(s, 130);
    while (n -- ){
        cin.getline(s, 130);
        // cout << s << "\n";
        int len = strlen(s);
        ans = 0;
        for (int i = 0; i < len - 1; i ++ ){
            if (s[i] == ' ')
                tmp1 = 0;
            else if (s[i] == '\'')
                tmp1 = 1;
            else
                tmp1 = int(s[i] - 'A') + 2;
            if (s[i + 1] == ' ')
                tmp2 = 0;
            else if (s[i + 1] == '\'')
                tmp2 = 1;
            else
                tmp2 = int(s[i + 1] - 'A') + 2;
            res = 60 * pi / 28.0 * min(abs(tmp1 - tmp2), 28 - abs(tmp1 - tmp2));
            ans += res / 15.0;
        }
        cout << fixed << setprecision(6) << ans + len << "\n";
    }
    return 0;
}

J. Lost Map

  • 思路：最小生成树 数组开小了段错误+1

  • AC代码

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll mult_mod(ll x, ll y, ll mod){
    return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}

ll pow_mod(ll a, ll b, ll p){
    ll res = 1;
    while (b){
        if (b & 1)
            res = mult_mod(res, a, p);
        a = mult_mod(a, a, p);
        b >>= 1;
    }
    return res % p;
}

ll gcd(ll a, ll b){
    return b ? gcd(b, a % b) : a;
}

const int N = 2510;

int n, tot;
int fa[N];
int mp[N][N];

void init(){
    for (int i = 1; i <= N; i ++ )
        fa[i] = i;
}

int find(int x){
    return x == fa[x] ? x : fa[x] = find(fa[x]);
}

void join(int x, int y){
    if (find(x) != find(y))
        fa[find(x)] = find(y);
}

struct Edge{
    int u, v, dis;
    Edge(){ };
    Edge(int _u, int _v, int _dis): u(_u), v(_v), dis(_dis){ };
    bool operator <(const Edge &e) const{
        return dis < e.dis;
    }
}e[N * N];

int main(){
#ifndef ONLINE_JUDGE
    freopen("my_in.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    init();
    cin >> n;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
            cin >> mp[i][j];
    for (int i = 1; i <= n; i ++ ){
        for (int j = i + 1; j <= n; j ++ ){
            Edge edge(i, j, mp[i][j]);
            e[tot ++ ] = edge;
        }
    }
    sort(e, e + tot);
    for (int i = 0; i < tot; i ++ ){
        if (find(e[i].u) == find(e[i].v))
            continue;
        join(e[i].u, e[i].v);
        cout << e[i].u << " " << e[i].v << "\n";
    }
    return 0;
}
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