2020牛客暑期多校训练营（第八场）

A All-Star Game
B Bon Voyage
C Cinema
D Disgusting Relationship
E Enigmatic Partition
F Factorio
G. Game SET

#include 
using namespace std;
const int N = 1e6 + 10;
int n;

string str[13] = {
        "one", "two", "three",
        "diamond", "squiggle", "oval",
        "solid", "striped", "open",
        "red", "green", "purple",
        "*"
};
map<string, int> mp;

bool check(int a, int b, int c) {
    if (a == 4) {
        if (b == 4) {
            // 就看c
            return true;
        } else {
            // b!=c 就构造a!=b!=c
            // b==c 就构造a==b==c
            return true;
        }
    } else {
        if (b == 4) {
            //构造a!=b!=c
            return true;
        } else {
            if (c == 4) {
                //构造a!=b!=c
                return true;
            } else {
                return (a == b && b == c) || (a != b && b != c && a != c);
            }
        }
    }
}

int a[N][4];

void getId(string s, int id) {
    for (int i = 0, j = 0, st, sz = s.length(); i < sz; i++) {
        if (s[i] == '[') {
            st = i + 1;
            while (s[++i] != ']');
            a[id][j++] = mp[s.substr(st, i - st)];
        }
    }
}

string s;

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

    mp[str[0]] = 1; mp[str[1]] = 2; mp[str[2]] = 3;
    mp[str[3]] = 1; mp[str[4]] = 2; mp[str[5]] = 3;
    mp[str[6]] = 1; mp[str[7]] = 2; mp[str[8]] = 3;
	mp[str[9]] = 1; mp[str[10]] = 2; mp[str[11]] = 3;
    mp[str[12]] = 4;

    int T;
    cin >> T;
    for (int cs = 1; cs <= T; cs++) {
        cout << "Case #" << cs << ": ";
        cin >> n;
        for (int i = 1; i <= n; i++) {
            cin >> s;
            getId(s, i);
        }
        
        int f1 = 0;
        // 显然 个数达到一定程度 肯定有解
        // 最多33^3 就能找到答案
        for (int i = 1, lim = min(33, n); i <= lim; i++) {
            for (int j = i + 1; j <= lim; j++) {
                for (int l = j + 1; l <= lim; l++) {
                    int f2 = 1;
                    for (int k = 0; k < 4; k++) {
                        if (!check(a[i][k], a[j][k], a[l][k])) {
                            f2 = 0;
                            break;
                        }
                    }

                    if (f2) {
                        f1 = 1;
                        cout << i << " " << j << " " << l << endl;
                        break;
                    }
                }
                if (f1) break;
            }
            if (f1) break;
        }

        if (!f1) {
            cout << -1 << endl;
        }
    }
    return 0;
}

H Hard String Problem

I. Interesting Computer Game

总结：$10$ 组 $3e5$ 的建图不适合网络流（要么趁出题人数据水时蒙个优化）

#include 

using namespace std;
const int N = 1e6 + 10;
int n;

namespace Discretization {
    vector<int> backUp;

    void discrete() {
        sort(backUp.begin(), backUp.end());
        backUp.erase(unique(backUp.begin(), backUp.end()), backUp.end());
    }

    int id(int x) {
        return lower_bound(backUp.begin(), backUp.end(), x) - backUp.begin() + 1;
    }
}
using namespace Discretization;


namespace Union_Find { // 并查集板子
    int sz[N], e_sz[N];
    int fa[N];

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

    void Union(int x, int y) {
        int fx = find(x);
        int fy = find(y);
        if (fx == fy) {
            e_sz[fx]++;
            return;
        }
        if (fx < fy) {
            fa[fy] = fx;
            sz[fx] += sz[fy];
            sz[fy] = 0;

            e_sz[fx] += e_sz[fy] + 1;
        } else {
            fa[fx] = fy;
            sz[fy] += sz[fx];
            sz[fx] = 0;
            e_sz[fy] += e_sz[fx] + 1;
        }
    }
}
using namespace Union_Find;

int a[N], b[N];

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

    int T;
    cin >> T;
    for (int cs = 1; cs <= T; cs++) {
        cout << "Case #" << cs << ": ";
        cin >> n;

        for (int i = 1; i <= n; i++) {
            cin >> a[i] >> b[i];
            backUp.push_back(a[i]);
            backUp.push_back(b[i]);
        }

        discrete();//离散化
        
        for (int i = 1, _sz = backUp.size(); i <= _sz; i++) {
            fa[i] = i;
            sz[i] = 1;
            e_sz[i] = 0;
        }

        for (int i = 1; i <= n; i++) {
            a[i] = id(a[i]);
            b[i] = id(b[i]);

            Union(a[i], b[i]);
        }

        int res = 0;
        for (int i = 1, _sz = backUp.size(); i <= _sz; i++) {
            if (i == find(i)) {
                res += min(sz[i], e_sz[i]);
            }
        }
        cout << res << endl;

        backUp.clear();
    }
    return 0;
}

J Jumping Points

K Kabaleo Lite

#include 
using namespace std;
typedef __int128 ll;
const int N = 1e5 + 10;

int a[N], b[N], Minb[N];
ll sum[N], Mx[N];
int n;

ll max(ll x, ll y) {
    return x >= y ? x : y;
}

void _print(ll x) {
    if (x > 9) _print(x / 10);
    putchar(x % 10 + '0');
}

void print(ll x) {//输出
    if (x < 0) {
        x = -x;
        putchar('-');
    }
    _print(x);
    puts("");
}

pair<ll, int> mp[N];
int tot = 0;

int main() {
    int T;
    scanf("%d", &T);
    for (int cs = 1; cs <= T; cs++) {
        printf("Case #%d: ", cs);
        scanf("%d", &n);

        tot = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }

        for (int i = 1; i <= n; i++) {
            scanf("%d", &b[i]);

            sum[i] = sum[i - 1] + a[i];

            if (i == 1) {
                Mx[i] = sum[i];
                Minb[i] = b[i];
            } else {
                Mx[i] = max(Mx[i - 1], sum[i]);
                Minb[i] = min(Minb[i - 1], b[i]);
            }

            if (Mx[i] == sum[i]) {
                mp[++tot] = {Mx[i], Minb[i]};
            }
        }

        int cnt = 0, x;//cnt 当前已经接待了多少游客
        ll res = 0;
        for (int i = tot; i > 0; i--) {
            x = max(0, mp[i].second - cnt);
            res += mp[i].first * x;
            cnt += x;
        }
        printf("%d ", b[1]);
        print(res);
    }
    return 0;
}