2017-2018 ACM-ICPC German Collegiate Programming Contest (GCPC 2017)（9/11）

$$2017-2018\ ACM-ICPC\ German\ Collegiate\ Programming\ Contest (GCPC 2017)$$

$A.Drawing\ Borders$

$B.Buildings$

Polya定理搞一搞

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
using LL = int_fast64_t;
const LL MOD = 1e9+7;
int n,m,c;
LL qpow(LL a, LL b){
    LL ret = 1;
    while(b){
        if(b&1) ret = ret * a % MOD;
        b >>= 1;
        a = a * a % MOD;
    }
    return ret;
}
LL inv(LL x){ return qpow(x,MOD-2); }
int main(){
    ____();
    cin >> n >> m >> c;
    LL ret = 0;
    for(int i = 0; i < m; i++) ret = (ret + qpow(qpow(c,n*n),__gcd(i,m))) % MOD;
    ret = (ret * inv(m)) % MOD;
    cout << ret << endl;
    return 0;
}

$C.Joyride$

DP,$dp[i][j]$表示当前时间为$i$，且当前在$j$号点处的最小花费，分两种情况转移

在原地不动，位置不变
去另一个位置

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1111;
int x,n,m,t,f[MAXN][MAXN];
const int INF = 0x3f3f3f3f;
pair<int,int> pr[MAXN];
vector<int> G[MAXN];
int main(){
    ____();
    cin >> x >> n >> m >> t;
    for(int i = 1; i <= m; i++){
        int u, v;
        cin >> u >> v;
        G[u].emplace_back(v);
        G[v].emplace_back(u);
    }
    for(int i = 1; i <= n; i++) cin >> pr[i].first >> pr[i].second;
    memset(f,0x3f,sizeof(f));
    if(x<pr[1].first){
        cout << "It is a trap." << endl;
        return 0;
    }
    f[pr[1].first][1] = pr[1].second;
    for(int c = pr[1].first; c < x; c++){
        for(int u = 1; u <= n; u++){
            if(c+pr[u].first<=x) f[c+pr[u].first][u] = min(f[c+pr[u].first][u],f[c][u]+pr[u].second);
            for(int v : G[u]){
                int cost = t + pr[v].first;
                if(c+cost>x) continue;
                f[c+cost][v] = min(f[c+cost][v],f[c][u]+pr[v].second);
            }
        }
    }
    int ret = f[x][1];
    if(ret==INF) cout << "It is a trap." << endl;
    else cout << ret << endl;
    return 0;
}

$D.Pants\ On\ Fire$

Floyd传递闭包

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 222;
map<string,int> msk;
int n,m,ID;
bool G[MAXN][MAXN];
int main(){
    ____();
    cin >> n >> m;
    for(int i = 1; i <= n; i++){
        int u,v;
        string s;
        cin >> s;
        if(!msk.count(s)) msk.insert(make_pair(s,++ID));
        u = msk.at(s);
        cin >> s >> s >> s >> s;
        if(!msk.count(s)) msk.insert(make_pair(s,++ID));
        v = msk.at(s);
        G[u][v] = true;
    }
    for(int k = 1; k <= ID; k++) for(int i = 1; i <= ID; i++) for(int j = 1; j <= ID; j++){
        G[i][j] = (G[i][j] or (G[i][k] and G[k][j]));
    }
    for(int i = 1; i <= m; i++){
        int u, v;
        string s;
        cin >> s;
        if(!msk.count(s)) msk.insert(make_pair(s,++ID));
        u = msk.at(s);
        cin >> s >> s >> s >> s;
        if(!msk.count(s)) msk.insert(make_pair(s,++ID));
        v = msk.at(s);
        if(G[u][v]) cout << "Fact" << endl;
        else if(G[v][u]) cout << "Alternative Fact" << endl;
        else cout << "Pants on Fire" << endl;
    }
    return 0;
}

$E.Perpetuum\ Mobile$

建图spfa判正环，可以对权值取$log$就可以转换为权值加和

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 888;
int n,m,vis[MAXN],inq[MAXN];
double dist[MAXN];
vector<pair<int,double> > G[MAXN];
bool spfa(){
    queue<int> que;
    memset(inq,0,sizeof(inq));
    fill(dist,dist+MAXN,0);
    fill(vis,vis+MAXN,1);
    for(int i = 1; i <= n; i++) que.push(i);
    while(!que.empty()){
        int u = que.front();
        que.pop();
        vis[u] = false;
        for(auto e : G[u]){
            int v = e.first;
            double d = e.second;
            if(dist[u]+d>dist[v]){
                dist[v] = dist[u] + d;
                if(!vis[v]){
                    vis[v] = true;
                    inq[v]++;
                    if(inq[v]>=n) return true;
                    que.push(v);
                }
            }
        }
    }
    return false;
}
int main(){
    cin >> n >> m;
    for(int i = 1; i <= m; i++) {
        int u, v;
        double conv;
        cin >> u >> v >> conv;
        G[u].emplace_back(make_pair(v,log(conv)));
    }
    if(spfa()) cout << "inadmissible" << endl;
    else cout << "admissible" << endl;
    return 0;
}

$F.Plug\ It\ In$

暴力枚举，每次跑二分图匹配，这样的复杂度是$n^3$，会T
考虑加上新的点之后，之前的匹配是不变的，所以可以先把之前的匹配保存下来，每次枚举的时候找增广路即可

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1507;
vector<int> G[MAXN];
int m,n,k,match[MAXN],tot,pmt[MAXN];
bool vis[MAXN];
bool dfs(int u){
    vis[u] = true;
    for(int v : G[u]){
        if(match[v]==-1 or (!vis[match[v]] and dfs(match[v]))){
            match[v] = u;
            return true;
        }
    }
    return false;
}
void hungary(){
    memset(match,255,sizeof(match));
    for(int i = 1; i <= m; i++){
        memset(vis,false,sizeof(vis));
        if(dfs(i)) tot++;
    }
}
int main(){
    ____();
    cin >> m >> n >> k;
    for(int i = 1; i <= k; i++){
        int u, v;
        cin >> u >> v;
        G[u].emplace_back(v);
    }
    hungary();
    for(int i = 1; i <= n; i++) pmt[i] = match[i];
    int add = 0;
    for(int i = 1; i <= m; i++){
        int tmp = 0;
        for(int j = 1; j <= n; j++) match[j] = pmt[j];
        G[m+1].clear();
        copy(G[i].begin(),G[i].end(),back_inserter(G[m+1]));
        memset(vis,false,sizeof(vis));
        if(dfs(m+1)){
            tmp++;
            G[m+2].clear();
            copy(G[m+1].begin(),G[m+1].end(),back_inserter(G[m+2]));
            memset(vis,false,sizeof(vis));
            if(dfs(m+2)) tmp++;
        }
        add = max(add,tmp);
    }
    cout << tot + add << endl;
    return 0;
}

$G.Water\ Testing$

皮克定理，顶点为整点的多边形的面积S=多边形内部整点数A+(多边形边上整点数B/2)-1
多边形面积可以用向量差积来求
所以$A=S+1-B/2$

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e5+7;
using LL = int_fast64_t;
int n;
pair<LL,LL> pt[MAXN];
pair<LL,LL> vect(const pair<LL,LL> &A, const pair<LL,LL> &B){
    return make_pair(B.first-A.first,B.second-A.second);
}
LL xpro(const pair<LL,LL> &A, const pair<LL,LL> &B){
    return A.first * B.second - B.first * A.second;
}
LL cal(const pair<LL,LL> &A, const pair<LL,LL> &B){
    return __gcd(abs(A.first-B.first),abs(A.second-B.second)) - 1;
}
int main(){
    ____();
    cin >> n;
    for(int i = 1; i <= n; i++) cin >> pt[i].first >> pt[i].second;
    LL dbarea = 0;
    for(int i = 2; i < n; i++) dbarea += xpro(vect(pt[1],pt[i]),vect(pt[1],pt[i+1]));
    dbarea = dbarea < 0 ? - dbarea : dbarea;
    LL tot = n;
    for(int i = 1; i < n; i++) tot += cal(pt[i],pt[i+1]);
    tot += cal(pt[1],pt[n]);
    cout << (dbarea-tot+2)/2 << endl;
    return 0;
}

$H.Ratatoskr$

考虑如何采取最佳决策，对于松鼠，最佳位置就是在两只乌鸦中间，这样不管乌鸦怎么跑，松鼠都可以从起飞的乌鸦的那一边走，对于乌鸦的最佳决策就是两只乌鸦交替向松鼠的方向移动，把松鼠逼到叶子节点即可，所以最差的情况，就是找一个根节点，然后把松鼠逼到叶子节点上，也就是找任意点为根的最大深度的最小值（这个情况下就是先堵住根节点，然后推进到叶子节点）。
还有一种情况就是，一只乌鸦先固定，然后往松鼠存在的那棵子树上逼近即可

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 88;
vector<int> G[MAXN];
int n,r,h,m,vis[MAXN];
int dfs(int u, int f, int d){
    int maxd = d; vis[u] = true;
    for(int v : G[u]) if(v!=f) maxd = max(maxd,dfs(v,u,d+1));
    return maxd;
}
int giao(int raven){
    memset(vis,0,sizeof(vis));
    for(int v : G[raven]){
        int dep = dfs(v,raven,1);
        if(vis[r]) return dep;
    }
}
int main(){
    ____();
    cin >> n >> r >> h >> m;
    for(int i = 1; i < n; i++){
        int u, v;
        cin >> u >> v;
        G[u].emplace_back(v);
        G[v].emplace_back(u);
    }
    int ret = MAXN;
    for(int i = 1; i <= n; i++) ret = min(ret,dfs(i,0,1));
    ret = min(ret,min(giao(h),giao(m)));
    cout << ret << endl;
    return 0;
}

$I.Uberwatch$

DP，$dp[i]$表示当前时刻为$i$的最多总分，状态转移方程为$dp[i]=max_{j=1}^{i-m}dp[j]+A[i]$

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 3e5+7;
int n,m,A[MAXN];
struct SegmentTree{
    int maxx[MAXN<<2],l[MAXN<<2],r[MAXN<<2];
    #define ls(rt) rt << 1
    #define rs(rt) rt << 1 | 1
    #define pushup(rt) maxx[rt] = max(maxx[ls(rt)],maxx[rs(rt)])
    void build(int L, int R, int rt){
        l[rt] = L; r[rt] = R;
        if(L+1==R) return;
        int mid = (L + R) >> 1;
        build(L,mid,ls(rt)); build(mid,R,rs(rt));
    }
    void update(int pos, int rt, int x){
        if(l[rt]+1==r[rt]){
            maxx[rt] = x;
            return;
        }
        int mid = (l[rt] + r[rt]) >> 1;
        if(pos<mid) update(pos,ls(rt),x);
        else update(pos,rs(rt),x);
        pushup(rt);
    }
    int qmax(int L, int R, int rt){
        if(l[rt]>=R or L>=r[rt]) return 0;
        if(L<=l[rt] and r[rt]<=R) return maxx[rt];
        return max(qmax(L,R,ls(rt)),qmax(L,R,rs(rt)));
    }
}ST;
int main(){
    ____();
    cin >> n >> m;
    for(int i = 1; i <= n; i++) cin >> A[i];
    ST.build(1,n+1,1);
    for(int t = m + 1; t <= n; t++) ST.update(t,1,ST.qmax(1,t-m+1,1)+A[t]);
    cout << ST.qmax(1,n+1,1) << endl;
    return 0;
}

$J.Word\ Clock$

$K.You Are Fired$

签到

//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e4+7;
pair<int,string> pr[MAXN];
int n,d,k;
int main(){
    ____();
    cin >> n >> d >> k;
    for(int i = 1; i <= n; i++) cin >> pr[i].second >> pr[i].first;
    sort(pr+1,pr+1+n,[](const pair<int,string> &A, const pair<int,string> &B){
        return A.first > B.first;
    });
    vector<string> vec;
    int tot = 0;
    for(int i = 1; i <= min(d,k); i++){
        tot += pr[i].first;
        vec.emplace_back(pr[i].second);
        if(tot>=d) break;
    }
    if(tot<d){
        cout << "impossible" << endl;
        return 0;
    }
    cout << vec.size() << endl;
    for(auto name : vec) cout << name << ", YOU ARE FIRED!" << endl;
    return 0;
}

posted @ 2020-03-18 19:59 _kiko 阅读(290) 评论(0) 编辑收藏举报

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_kiko

2017-2018 ACM-ICPC German Collegiate Programming Contest (GCPC 2017)（9/11）

$$2017-2018\ ACM-ICPC\ German\ Collegiate\ Programming\ Contest (GCPC 2017)$$

\(A.Drawing\ Borders\)

\(B.Buildings\)

\(C.Joyride\)

\(D.Pants\ On\ Fire\)

\(E.Perpetuum\ Mobile\)

\(F.Plug\ It\ In\)

\(G.Water\ Testing\)

\(H.Ratatoskr\)

\(I.Uberwatch\)

\(J.Word\ Clock\)

\(K.You Are Fired\)

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