2021 ICPC North American Qualifier I. Pizza Party!(暴力)

这个题，属实有点唬人，学妹问我是不是拓扑排序

就是给了一些已经有的材料，和一些and和or的条件，只有满足前缀才能获得后一种材料，问最后能有几种材料

我一开始想的是网络流带上下界，然后看了下过题人数感觉有点问题，因为那么复杂的建图不可能那么多人都掌握

结果看到数据只有1k，不知道为什么想到了bellman ford和之前的leetcode杯的那个最短路

然后忽然想到了，因为暴力模拟无论输入的条件是什么顺序，我们都可以在n次之后到达稳定状态

那就扔进set暴力模拟一下就完事儿了

也不知道Luka为什么会在这题上面翻车23333

#include <bits/stdc++.h>

using namespace std;
#define limit (1000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;

inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}


const ll mod = 1e9 + 7;

ll quickPow(ll base, ll expo) {
    ll ans = 1;
    while (expo) {
        if (expo & 1)(ans *= base) %= mod;
        expo >>= 1;
        base = base * base;
        base %= mod;
    }
    return ans % mod;
}

ll C(ll n, ll m) {
    if (n < m)return 0;
    ll x = 1, y = 1;
    if (m > n - m)m = n - m;
    rep(i, 0, m - 1) {
        x = x * (n - i) % mod;
        y = y * (i + 1) % mod;
    }
    return x * quickPow(y, mod - 2) % mod;
}

ll lucas(ll n, ll m) {
    return !m || n == m ? 1 : C(n % mod, m % mod) * lucas(n / mod, m / mod) % mod;
}
ll fact[limit];
void calc(){
    fact[0] = 1;
    rep(i,1,1e3)fact[i] = (fact[i - 1] * i) % mod;
}
ll A(ll x, ll y){
    return (C(x,y) * fact[y]) % mod;
}
int kase;
int n, m, k;
int a[limit];
int tot;
map<string, int>mp;
int id(string x){
    auto & it = mp[x];
    if(!it)it = ++tot;
    return it;
}
set<pi(pi(int, int), pi(int, int))>constraint;
void solve() {
    cin>>n;
    set<int>s;
    rep(i,1,n){
        string str;
        cin>>str;
        if(str != "if"){
            s.insert(id(str));
        }else{
            string token;
            cin>>token;
            int fst = id(token);
            cin>>token;
            string expr = token;
            cin>>token;
            int scd = id(token);
            cin>>token;
            cin>>token;
            int cur_id = id(token);
            if(expr == "and"){
                if(fst > scd)swap(fst, scd);
                constraint.insert({{fst, scd}, {cur_id, 0}});
            }else{
                if(fst > scd)swap(fst, scd);
                constraint.insert({{fst, scd}, {cur_id, 1}});
            }
        }
    }
    int last = -1;
    int cur = s.size();
    rep(i,1,n<<1){
        vector<pi(pi(int, int), pi(int, int))>erased;
        for(auto it : constraint){
            auto [u, oper] = it;
            auto [key, op] = oper;
            auto [fst, scd] = u;
            if(!op){
                if(s.count(fst) and s.count(scd)){
                    s.insert(key);
                    erased.push_back(it);
                }
            }else{
                if(s.count(fst) or s.count(scd)){
                    s.insert(key);
                    erased.push_back(it);
                }
            }

        }
        for(auto it: erased)constraint.erase(it);
    }
    cout<<s.size()<<endl;
}

int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
//    cin >> kase;
//    while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
    return 0;
}

标签：North,Qualifier,int,scd,ll,ICPC,token,mod,define
来源： https://www.cnblogs.com/tiany7/p/15842586.html