ACM-ICPC 2018 南京赛区网络预赛

A An Olympian Math Problem

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll T,n;
int main(){
    scanf("%lld",&T);
    while(T--){
        scanf("%lld",&n);
        printf("%lld\n",n-1);
    }
}

B The writing on the wall

C GDY

D Jerome‘s House

E AC Challenge

#include <bits/stdc++.h>
using namespace std;
long long a[25],b[25];
long long dis[2000005];
int vis[2000005];
int pre[25];
int sta[25];
int main()
{
        memset(sta,0,sizeof(sta));
        memset(dis,0,sizeof(dis));
        memset(vis,0,sizeof(vis));
        int n;
        scanf("%d",&n);

        for(int i = 0;i < n;++i){
            scanf("%lld%lld%d",&a[i],&b[i],&pre[i]);
            int p;
            for(int j = 0;j < pre[i];++j){
                scanf("%d",&p);
                sta[i] |= (1<<(p-1));
            }
            //cout << sta[i] << endl;
        }
        int now = 0;
        queue<int> Q;
        while(!Q.empty()) Q.pop();
        Q.push(0);
        while(!Q.empty()){
            now = Q.front();
            Q.pop();
            vis[now] = 0;
            for(int i = 0;i < n;++i){
                if((sta[i] & now) == sta[i] && ((1<<i) & now) == 0){
                    int nxt = now | (1<<i);
                    int cnt = 0;
                    for(int j = 0;j < n;++j) if((1<<j)&nxt) cnt++;
                    if(dis[nxt] < dis[now] + cnt*a[i] + b[i]){
                        dis[nxt] = dis[now] + cnt*a[i] + b[i];
                        if(!vis[nxt]) Q.push(nxt),vis[nxt] = 1;
                    }
                }
            }
        }
        int up = (1<<n);
        long long ans = 0;
        for(int i = 0;i < up;++i) ans = max(ans,dis[i]);
        cout << ans << endl;
        return 0;
}

F An Easy Problem On The Trees

G Lpl and Energy-saving Lamps

H Set

I Skr

J Sum

#include <bits/stdc++.h>
using namespace std;
#define N 20000006
int prime[N],tag[N];
int get_prime(int n){
    int tot = 0;
    for(int i = 2;i <= n;++i){
        if(!tag[i]) prime[tot++] = i;
        for(int j = 0;j < tot && 1LL * prime[j] * i <= n;++j){
            tag[i * prime[j]] = 1;
            if(i % prime[j] == 0) break;
        }
    }
    return tot;
}

int cnt;
int ct;
int dfs(int now,int n){
    ct++;
    if(n == 0) return 0;
    if(now == cnt || prime[now] > n) return 1;
    if(prime[now]*prime[now] > n){
        int tmp = (upper_bound(prime,prime+cnt,n)-prime)-now;
        return tmp*2+1;
    }
    int ans = 0;
    ans += dfs(now+1,n/(1LL*prime[now]*prime[now]));
    ans += 2*dfs(now+1,n/prime[now]);
    ans += dfs(now+1,n);
    return ans;
}

int main()
{
    cnt = get_prime(20000000);
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        ct = 0;
        printf("%d\n",dfs(0,n));
        //cout << ct << endl;
    }
    return 0;
}

K The Great Nim Game

L Magical Girl Haze

#include <bits/stdc++.h>
using namespace std;
#define M 200005
#define N 100005
int head[N];
bool vis[N][11];
long long dis[N][11];
int cnt;

struct Edge{
    int to,next,l;
}edge[M<<1];

void ad(int u,int v,int c){
    edge[cnt].to = v,edge[cnt].next = head[u],edge[cnt].l = c,head[u] = cnt++;
}
struct Node{
    int p,tim;
    long long dis;
    bool operator < (const Node &a)const{
        return dis > a.dis;
    }
};
priority_queue<Node> Q;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        cnt = 0;
        memset(head,-1,sizeof(head));
        memset(dis,0x3f,sizeof(dis));
        memset(vis,0,sizeof(vis));
        int n,m,k;
        int u,v,c;
        scanf("%d%d%d",&n,&m,&k);
        for(int i = 1;i <= m;++i){
            scanf("%d%d%d",&u,&v,&c);
            ad(u,v,c);
        }
        dis[1][0] = 0;
        while(!Q.empty()) Q.pop();
        Node nd;
        nd.p = 1,nd.tim = 0,nd.dis = dis[1][0];
        Q.push(nd);
        long long l;
        while(!Q.empty()){
            nd = Q.top();
            Q.pop();
            int u = nd.p,tim = nd.tim;
            if (vis[u][tim]) continue;
            vis[u][tim] = 1;
            for(int i = head[u];i != -1;i = edge[i].next){
                v = edge[i].to;
                l = edge[i].l;
                if(tim + 1 <= k){
                    if(!vis[v][tim+1] && dis[v][tim+1] > dis[u][tim]){
                        dis[v][tim+1] = dis[u][tim];
                        nd.p = v,nd.tim = tim+1,nd.dis = dis[v][tim+1];
                        Q.push(nd);
                    }
                }
                if(!vis[v][tim] && dis[v][tim] > dis[u][tim] + l){
                    dis[v][tim] = dis[u][tim] + l;
                    nd.p = v,nd.tim = tim,nd.dis = dis[v][tim];
                    Q.push(nd);
                }
            }
        }
        long long ans = 0x3f3f3f3f3f3f3f3f;
        for(int i = 0;i <= k;++i) ans = min(ans,dis[n][i]);
        cout << ans << endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/solvit/p/9651897.html