2015ACM/ICPC亚洲区沈阳站 Solution

A - Pattern String

留坑。

B - Bazinga

题意：找一个最大的i，使得前i - 1个字符串中至少不是它的子串

思路：暴力找，如果有一个串已经符合条件，就不用往上更新

#include <bits/stdc++.h>
using namespace std;

#define N 510

int t, n;
int vis[N];
string s[N];

int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    for(int cas = 1; cas <= t; ++cas)
    {
        cin >> n;
        memset(vis, 0, sizeof vis);
        for(int i = 1; i <= n; ++i) cin >> s[i];
        int ans = -1;
        for(int i = 1; i <= n; ++i)
        {
            for(int j  = i + 1; j <= n; ++j)
            {
                if(!vis[j])
                {
                    int tmp = s[j].find(s[i]);
                    if(tmp == -1)
                    {
                        vis[j] = 1;
                        ans = max(ans, j);
                    }
                    else break;
                }
            }
        }
        cout << "Case #" << cas << ": " << ans << endl;
    }
    return 0;
}

C - Minimum Cut-Cut

留坑。

D - Pagodas

题意：刚开始集合里有a, b 两个数，每次能从集合中选出 x + y  或者 x - y 放进集合中，并且放过的不能放，不能放的输游戏

思路：考虑$gcd(x, y), 每次产生的新数肯定是gcd(x, y)的倍数$ 所以能放的数总数只有 gcd(x, y) 的倍数的个数

#include<bits/stdc++.h>

using namespace std;

int gcd(int a, int b)
{
    return b == 0 ? a : gcd(b, a % b);
}

int n, a, b;

int main()
{
    int t;
    scanf("%d", &t);
    for(int cas = 1; cas <= t; ++cas)
    {
        scanf("%d %d %d", &n, &a, &b);
        int g = gcd(a, b);
        int tmp = n / g;
        printf("Case #%d: ", cas);
        puts(tmp & 1 ? "Yuwgna" : "Iaka");
    }
    return 0;
}

E - Efficient Tree

留坑。

F - Frogs

题意：有n只青蛙，长度为m的环，每只青蛙固定步数往前跳，跳到一个点标记一下，求最后有多少点被标记

思路：先将m分解因子，对于每一个arr[i]都和m求gcd，然后将gcd的倍数标记一次，代表需要统计一次它的倍数。最后遍历一边每个因子，求出它被多用几次或者少用几次的和即可。

#include<bits/stdc++.h>

using namespace std;

#define ll long long
#define N 10010

ll gcd(ll a, ll b)
{
    return b == 0 ?  a :  gcd(b, a % b);
}

int n, m;
int vis[N], used[N];
ll arr[N];
vector<int>vec;

int main()
{
    int t;
    scanf("%d", &t);
    for(int cas = 1; cas <= t; ++cas)
    {
        scanf("%d %d", &n, &m);
        memset(vis, 0, sizeof vis);
        memset(used, 0, sizeof used);
        vec.clear();
        for(int i = 1; i * i <= m; ++i)
        {
            if(m % i == 0)
            {
                vec.push_back(i);
                if(i * i != m) vec.push_back(m / i);
            }
        }
        sort(vec.begin(), vec.end());
        int tot = vec.size();
        tot--;
        for(int i = 1; i <= n; ++i)
        {
            scanf("%lld", arr + i);
            ll x = gcd(arr[i], m);
            for(int j = 0; j < tot; ++j)
            {
                if(vec[j] % x == 0) vis[j] = 1;
            }
        }
        ll ans = 0;
        for(int i = 0; i < tot; ++i)
        {
            if(vis[i] != used[i])
            {
                ll tmp = (m - 1) / vec[i];
                ans += tmp * (tmp + 1) / 2 * vec[i] * (vis[i] - used[i]);
                for(int j = i + 1; j < tot; ++j)
                {
                    if(vec[j] % vec[i] == 0) 
                    {
                        used[j] += vis[i] - used[i];
                    }                        
                }
            }
        }
        printf("Case #%d: %lld
", cas, ans);
    }
    return 0;
}

G - Game of Flying Circus

题意：翻译啊翻译

思路：分类讨论

1)在1-2相遇   即v1=v2  输出Yes

2)在2-3相遇，利用二分计算出相遇点，算一下到4的时间，判断

3)在3-4相遇，利用二分计算出相遇点，算一下到1的时间，判断

4)一定No

#include<bits/stdc++.h>

using namespace std;

const double eps = 1e-8;

double T, v1, v2;

int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    else return x > 0 ? 1 : -1;
}

int main()
{
    int t;
    scanf("%d", &t);
    for(int cas = 1; cas <= t; ++cas)
    {
        scanf("%lf %lf %lf", &T, &v1, &v2);
        printf("Case #%d: ", cas);
        if(sgn(v1 - v2) == 0) puts("Yes");
        else if(sgn(2 * v1 * v1 - v2 * v2) > 0)//2-3
        {
            double l = 0, r = 300.0;
            while(r - l > eps)
            {
                double mid = (l + r) / 2.0;
                double dis1 = sqrt(mid * mid + 300.0 * 300.0);
                double t1 = dis1 / v1;
                double dis2 = mid + 300.0;
                double t2 = dis2 / v2;
                if(sgn(t1 - t2) > 0)
                {
                    l = mid;
                }
                else 
                {
                    r = mid;
                }
            }
            double x = (l + r) / 2.0;
            double dis1 = x + 600.0;
            double t1 = dis1 / v1;
            double dis2 = 600.0 - x;
            double t2 = dis2 / v2 + T;
            puts(sgn(t1 - t2) <= 0 ? "Yes" : "No");
        }
        else if(sgn(3 * v1 - v2) > 0)//3-4
        {
            double l = 0, r = 300.0;
            while(r - l > eps)
            {
                double mid = (l + r) / 2.0;
                double dis1 = sqrt(mid * mid + 300.0 * 300.0);
                double t1 = dis1 / v1;
                double dis2 = 900.0 - mid;
                double t2 = dis2 / v2;
                if(sgn(t1 - t2) > 0)
                {
                    r = mid;
                }
                else 
                {
                    l = mid;
                }
            }
            double y = (l + r) / 2.0;
            double dis1 = sqrt((300.0 - y) * (300.0 - y) + 300.0 * 300.0) + 900.0;
            double t1 = dis1 / v1;
            double dis2 = y + 300.0;
            double t2 = dis2 / v2 + T;
            puts(sgn(t1 - t2) <= 0 ? "Yes" : "No");
        }
        else puts("No");
    }
    return 0;
}

H - Chessboard

留坑。

I - Triple

题意：二元组$A<a, b>$ 三元组 $B<c, d, e> A * B = {<a, c, d> | <a, b> in A , <c, d, e> in B and b = e}$ 求有多少集合满足TOP(C)

思路：枚举B，对于每个B，我们考虑符合条件的最高的a, 因为有多个a，那么小的a肯定不会放到集合里面，

对于最大的a，我们判断二维BIT中右下角有的最大值，是否大于它本身，不是的话它就应该放进TOP(C) 

要多考虑一下当前的(c, d) 有多个点的情况

#include <bits/stdc++.h>
using namespace std;

#define N 100010
#define pii pair <int, int>
#define ll long long

int t, n, m;
struct A
{
    int a, b;
    void scan()
    {
        scanf("%d%d", &a, &b);
    }
    bool operator < (const A &r) const
    {
        return b == r.b ? a < r.a : b < r.b;
    }
}a[N];

struct B
{
    int c, d, e;
    void scan()
    {
        scanf("%d%d%d", &c, &d, &e);
    }
}b[N];

pii pos[N];  

struct BIT
{
    int a[1010][1010];

    void Init()
    {
        memset(a, 0, sizeof a);
    }

    void update(int x, int y, int val)
    {
        for (int i = x; i; i -= i & -i)
            for (int j = y; j; j -= j & -j)
                a[i][j] = max(a[i][j], val);
    }

    int query(int x, int y)
    {
        int res = 0;
        for (int i = x; i < 1010; i += i & -i)
            for (int j = y; j < 1010; j += j & -j)
                res = max(res, a[i][j]);
        return res;
    }
}bit; 

int main()
{
    scanf("%d", &t);
    for (int kase = 1; kase <= t; ++kase)
    {
        printf("Case #%d: ", kase); 
        scanf("%d%d", &n, &m); 
        for (int i = 1; i <= n; ++i) a[i].scan(); 
        for (int i = 1; i <= m; ++i) b[i].scan(); 
        sort(a + 1, a + 1 + n);     
        memset(pos, 0, sizeof pos); 
        for (int i = n; i >= 1; --i)
        {
            int b = a[i].b;
            if (!pos[b].first)
            {
                pos[b].first = a[i].a, ++pos[b].second; 
                continue;    
            }
            if (i != n) 
            {
                if (b == a[i + 1].b && a[i].a == pos[b].first) 
                    ++pos[b].second; 
            }
        }
        bit.Init(); 
        for (int i = 1; i <= m; ++i) 
        {
            if (pos[b[i].e].first == 0) continue;
            bit.update(b[i].c, b[i].d, pos[b[i].e].first);
        }
        ll res = 0; 
        for (int i = 1; i <= m; ++i)
        {
            int x = b[i].c, y = b[i].d;
            if(!pos[b[i].e].first) continue;
            int Max = max(bit.query(x, y + 1), bit.query(x + 1, y));
            bool flag = true; 
            if (Max >= pos[b[i].e].first) flag = false;
            Max = bit.query(x, y); if (Max != pos[b[i].e].first) flag = false;
            if (flag) res += pos[b[i].e].second;
        }
        printf("%lld
", res);
    }
    return 0;
}

J - John's Fences

留坑。

K - Kykneion asma

留坑。

L - Number Link

留坑。

M - Meeting

题意：有多个集合，集合里两个点的距离都是$t-i$ 1 和 n  要到一个地方汇合，使得最大距离最小，求汇合点

思路：考虑暴力连边的边数最大有${sum_{i = 1}^{i = m} S_i} ^2$ 没法解决

考虑将一个集合里所有点向另一个点连权值为$t_i$的边，再从这个点向集合里所有点连权值为0的边

再对1和n分别跑最短路，枚举中间点，更新答案

#include <bits/stdc++.h>
using namespace std;

#define N 2000010
#define ll long long
#define INFLL 0x3f3f3f3f3f3f3f3f

int t, n, m;
int arr[N];

struct Edge
{
    int to, nx; ll w;
    Edge() {}
    Edge(int to, int nx, ll w) : to(to), nx(nx), w(w) {}
}edge[N << 1]; 

int head[N], pos;

void Init()
{
    memset(head, -1, sizeof head);
    pos = 0;
}

void addedge(int u, int v, int w)
{
    edge[++pos] = Edge(v, head[u], w); head[u] = pos;
}

struct node
{
    int u; ll w;
    node () {}
    node (int u, ll w) : u(u), w(w) {}
    bool operator < (const node &r) const
    {
        return w > r.w;
    }
};

ll dist[2][N];
bool used[N];

void Dijkstra(int st, int id)
{
    for (int i = 1; i <= n + m; ++i) dist[id][i] = INFLL, used[i] = 0;
    priority_queue <node> q;
    q.emplace(st, 0);
    dist[id][st] = 0;
    while (!q.empty())
    {
        int u = q.top().u; q.pop();
        if (used[u]) continue;
        used[u] = 1;
        for (int it = head[u]; ~it; it = edge[it].nx)
        {
            int v = edge[it].to;
            if (!used[v] && dist[id][v] > dist[id][u] + edge[it].w)
            {
                dist[id][v] = dist[id][u] + edge[it].w;
                q.emplace(v, dist[id][v]);
            }
        }
    }
}


int main()
{
    scanf("%d", &t);
    for (int kase = 1; kase <= t; ++kase)
    {
        printf("Case #%d: ", kase);
        scanf("%d%d", &n, &m);
        Init();
        for (int i = 1, tot, w; i <= m; ++i)
        {
            scanf("%d%d", &w, &tot);
            for (int j = 1; j <= tot; ++j) scanf("%d", arr + j);
            for (int j = 1; j <= tot; ++j) addedge(arr[j], i + n, w);
            for (int j = 1; j <= tot; ++j) addedge(i + n, arr[j], 0); 
        }    
        Dijkstra(1, 0);
        Dijkstra(n, 1);
        vector <int> v;
        ll Min = INFLL;
//        for (int i = 1; i <= n; ++i) cout << dist[0][i] << " " << dist[1][i] << endl;
        for (int i = 1; i <= n; ++i) Min = min(Min, max(dist[0][i], dist[1][i]));
        if (Min == INFLL) puts("Evil John");
        else
        {
            for (int i = 1; i <= n; ++i) if (max(dist[0][i], dist[1][i]) == Min)
                v.push_back(i);
            printf("%lld
", Min);
            for (int i = 0, len = v.size(); i < len; ++i) printf("%d%c", v[i], " 
"[i == len - 1]);
        }
    }
    return 0;
}

C=A∗B={⟨a,c,d⟩∣⟨a,b⟩∈A, ⟨c,d,e⟩∈B and b=e}