A Drawing Borders

很多构造方法，下图可能是最简单的了

代码：

#include<bits/stdc++.h>

using namespace std;

const int maxn=1e6+;

struct Point{ int x,y; };

Point a[maxn];  int numa=;

Point b[maxn];  int numb=;

vector<pair<double,double> > va;

vector<pair<double,double> > vb;

void checka()

{

    va.push_back({,});

    va.push_back({,});

    for(int i=numa;i>=;i--)

    {

        va.push_back( { a[i].x*1.000-0.3,} );

        va.push_back( { a[i].x*1.000-0.3,a[i].y*1.000+0.1} );

        va.push_back( { a[i].x*1.000+0.1,a[i].y*1.000+0.1} );

        va.push_back( { a[i].x*1.000+0.1,a[i].y*1.000-0.1} );

        va.push_back( { a[i].x*1.000-0.3,a[i].y*1.000-0.1} );

        while(i->= && a[i-].x==a[i].x)

        {

            i--;

            va.push_back( { a[i].x*1.000-0.3,a[i].y*1.000+0.1} );

            va.push_back( { a[i].x*1.000+0.1,a[i].y*1.000+0.1} );

            va.push_back( { a[i].x*1.000+0.1,a[i].y*1.000-0.1} );

            va.push_back( { a[i].x*1.000-0.3,a[i].y*1.000-0.1} );

        }

        va.push_back( {a[i].x*1.0000-0.3,-} );

        va.push_back( {a[i].x*1.0000-0.39,-} );

        va.push_back( {a[i].x*1.0000-0.39,});

    }

    va.push_back({-,});

    va.push_back({-,});

}

void checkb()

{

    vb.push_back({-,-});

    vb.push_back({-,-});

    for(int i=;i<=numb;i++)

    {

        vb.push_back( { b[i].x*1.000+0.3,-} );

        vb.push_back( { b[i].x*1.000+0.3,b[i].y*1.000-0.1} );

        vb.push_back( { b[i].x*1.000-0.1,b[i].y*1.000-0.1} );

        vb.push_back( { b[i].x*1.000-0.1,b[i].y*1.000+0.1} );

        vb.push_back( { b[i].x*1.000+0.3,b[i].y*1.000+0.1} );

        while(i+<=numb && b[i+].x==b[i].x)

        {

            i++;

            vb.push_back( { b[i].x*1.000+0.3,b[i].y*1.000-0.1} );

            vb.push_back( { b[i].x*1.000-0.1,b[i].y*1.000-0.1} );

            vb.push_back( { b[i].x*1.000-0.1,b[i].y*1.000+0.1} );

            vb.push_back( { b[i].x*1.000+0.3,b[i].y*1.000+0.1} );

        }

        vb.push_back( { b[i].x*1.000+0.3,} );

        vb.push_back( { b[i].x*1.000+0.39,} );

        vb.push_back( { b[i].x*1.000+0.39,-} );

    }

    vb.push_back({,-});

    vb.push_back({,-});

}

bool upa(Point a,Point b){if(a.x==b.x) return a.y<b.y;return   a.x<b.x;}

bool upb(Point a,Point b){if(a.x==b.x) return a.y<b.y;return   a.x<b.x;}

int main()

{

    int n ; scanf("%d",&n);

    for(int i=;i<=n;i++)

    {

        int x,y,z; scanf("%d %d %d",&x,&y,&z);

        if(z==) a[++numa].x=x,a[numa].y=y;

        if(z==) b[++numb].x=x,b[numb].y=y;

    }

    sort(a+,a++numa,upa);   checka();

    sort(b+,b++numb,upb);   checkb();

    printf("%d\n",va.size()); for(int i=;i<va.size();i++) printf("%.4f %.4f\n",va[i].first,va[i].second);

    printf("%d\n",vb.size()); for(int i=;i<vb.size();i++) printf("%.4f %.4f\n",vb[i].first,vb[i].second);

}

B Buildings

polya定理

代码：

#include<bits/stdc++.h>

#define int long long

using namespace std;

const int mod=1e9+;

int quick(int a,int n)

{

    int ans=;

    while(n)

    {

        if(n%==) {a=a%mod*a%mod;n=n/; }

        else       {ans=ans%mod*a%mod;   n--;   }

    }

    return ans;

}

int p[+];

int32_t main()

{

   int n,m,c; cin>>n>>m>>c;//  m  temp;

   int temp=quick(c,n*n); //cout<<temp<<endl;

   p[]=;

   for(int i=;i<= m;i++) p[i]=p[i-]%mod*temp%mod;

   int a=;

   for(int i=;i<m;i++)  a+=p[__gcd(i,m)],a=a%mod;

   int ans=a*quick(m,mod-)%mod;

   cout<<ans<<endl;

}

C Joyride

dp + 记忆化搜索

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 1e3 + ;

const LL INF = 0x3f3f3f3f3f3f3f3f;

LL dp[N][N];

vector<int> g[N];

int x, n, m, T, u, v;

int t[N], p[N];

LL dfs(int u, int m) {

    if(m > x) return INF;

    if(~dp[u][m]) return dp[u][m];

    if(u ==  && m == x) return dp[u][m] = ;

    dp[u][m] = INF;

    for (int v : g[u]) {

        dp[u][m] = min(dp[u][m], dfs(v, m+T+t[v]) + p[v]);

    }

    dp[u][m] = min(dp[u][m], dfs(u, m+t[u])+p[u]);

    return dp[u][m];

}

int main() {

    mem(dp, -);

    scanf("%d", &x);

    scanf("%d %d %d", &n, &m, &T);

    for (int i = ; i <= m; ++i) scanf("%d %d", &u, &v), g[u].pb(v), g[v].pb(u);

    for (int i = ; i <= n; ++i) scanf("%d %d", &t[i], &p[i]);

    LL tmp = dfs(, t[]) + p[];

    if(t[]* > x || tmp >= INF) printf("It is a trap.\n");

    else printf("%lld\n", tmp);

    return ;

}

D Pants On Fire

拓扑排序

代码：

#include <algorithm>

#include  <iterator>

#include  <iostream>

#include   <cstring>

#include   <cstdlib>

#include   <iomanip>

#include    <bitset>

#include    <cctype>

#include    <cstdio>

#include    <string>

#include    <vector>

#include     <stack>

#include     <cmath>

#include     <queue>

#include      <list>

#include       <map>

#include       <set>

#include   <cassert>

/*

⊂_ヽ

　 ＼＼ Λ＿Λ  来了老弟

　　 ＼('ㅅ')

　　　 >　⌒ヽ

　　　/ 　 へ＼

　　 /　　/　＼＼

　　 ﾚ　ノ　　 ヽ_つ

　　/　/

　 /　/|

　(　(ヽ

　|　|、＼

　| 丿 ＼ ⌒)

　| |　　) /

'ノ )　　Lﾉ

*/

using namespace std;

#define lson (l , mid , rt << 1)

#define rson (mid + 1 , r , rt << 1 | 1)

#define debug(x) cerr << #x << " = " << x << "\n";

#define pb push_back

#define pq priority_queue

typedef long long ll;

typedef unsigned long long ull;

//typedef __int128 bll;

typedef pair<ll ,ll > pll;

typedef pair<int ,int > pii;

typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q

//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q

#define fi first

#define se second

//#define endl '\n'

#define boost ios::sync_with_stdio(false);cin.tie(0)

#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)

#define max3(a,b,c) max(max(a,b), c);

#define min3(a,b,c) min(min(a,b), c);

const ll oo = 1ll<<;

const ll mos = 0x7FFFFFFF;  //

const ll nmos = 0x80000000;  //-2147483648

const int inf = 0x3f3f3f3f;

const ll inff = 0x3f3f3f3f3f3f3f3f; //

const int mod = 1e9+;

const double esp = 1e-;

const double PI=acos(-1.0);

const double PHI=0.61803399;    //黄金分割点

const double tPHI=0.38196601;

template<typename T>

inline T read(T&x){

    x=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

inline void cmax(int &x,int y){if(x<y)x=y;}

inline void cmax(ll &x,ll y){if(x<y)x=y;}

inline void cmin(int &x,int y){if(x>y)x=y;}

inline void cmin(ll &x,ll y){if(x>y)x=y;}

/*-----------------------showtime----------------------*/

            string str[];

            map<string ,int>mp;

            vector<int>G[];

            int fa[],du[],dig[];

            int find(int x) {

                if(fa[x] == x) return x;

                return fa[x] = find(fa[x]);

            }

            int dfs(int s,int t){

                int flag = ;

                for(int i=; i<G[s].size(); i++){

                    int v = G[s][i];

                    if(v == t) flag = ;

                    flag = (flag || dfs(v, t));

                }

                return flag;

            }

int main(){

            int n,m;

            scanf("%d%d", &n, &m);

            int tot = ;

            rep(i, , ) fa[i] = i;

            rep(i, , n) {

                rep(j, , ) cin>>str[j];

                int u,v;

                if(mp.count(str[]))  u = mp[str[]];

                else {

                    ++tot; u = tot;

                    mp[str[]] = tot;

                }

                if(mp.count(str[]))  v = mp[str[]];

                else {

                    ++tot; v = tot;

                    mp[str[]] = tot;

                }

                int fx = find(u), fy = find(v);

                fa[fx]=  fy;

                G[u].pb(v);

            }

            for(int i=; i<=m; i++){

                 rep(j, , ) cin>>str[j];

                 if(mp.count(str[]) ==  || mp.count(str[]) == ) {

                    puts("Pants on Fire");

                    continue;

                 }

                 int u = mp[str[]];

                 int v = mp[str[]];

                 if(find(u) != find(v) ) puts("Pants on Fire");

                 else if(dfs(u, v)) puts("Fact");

                 else if(dfs(v, u))puts("Alternative Fact");

                 else puts("Pants on Fire");

            }

            return ;

}

E Perpetuum Mobile

spfa判正权环

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N =  + ;

double d[N];

struct edge {

    int to;

    double w;

};

vector<edge> g[N];

bool vis[N];

bool spfa(int u) {

    vis[u] = true;

    for (edge v : g[u]) {

        if(d[u] + v.w < d[v.to]) {

            d[v.to] = d[u] + v.w;

            if(vis[v.to]) return true;

            if(spfa(v.to)) return true;

        }

    }

    vis[u] = false;

    return false;

}

int main() {

    int n, m, u, v;

    double w;

    scanf("%d %d", &n, &m);

    for (int i = ; i <= m; ++i) {

        scanf("%d %d %lf", &u, &v, &w);

        g[u].pb({v, -log(w)});

    }

    for (int i = ; i <= n; ++i) d[i] = ;

    for (int i = ; i <= n; ++i) if(spfa(i)) {

        puts("inadmissible");

        return ;

    }

    puts("admissible");

    return ;

}

F Plug It In

枚举三个插座的位置，拆点，然后二分图匹配

代码：

#include <algorithm>

#include  <iterator>

#include  <iostream>

#include   <cstring>

#include   <cstdlib>

#include   <iomanip>

#include    <bitset>

#include    <cctype>

#include    <cstdio>

#include    <string>

#include    <vector>

#include     <stack>

#include     <cmath>

#include     <queue>

#include      <list>

#include       <map>

#include       <set>

#include   <cassert>

/*

⊂_ヽ

　 ＼＼ Λ＿Λ  来了老弟

　　 ＼('ㅅ')

　　　 >　⌒ヽ

　　　/ 　 へ＼

　　 /　　/　＼＼

　　 ﾚ　ノ　　 ヽ_つ

　　/　/

　 /　/|

　(　(ヽ

　|　|、＼

　| 丿 ＼ ⌒)

　| |　　) /

'ノ )　　Lﾉ

*/

using namespace std;

#define lson (l , mid , rt << 1)

#define rson (mid + 1 , r , rt << 1 | 1)

#define debug(x) cerr << #x << " = " << x << "\n";

#define pb push_back

#define pq priority_queue

typedef long long ll;

typedef unsigned long long ull;

//typedef __int128 bll;

typedef pair<ll ,ll > pll;

typedef pair<int ,int > pii;

typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q

//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q

#define fi first

#define se second

//#define endl '\n'

#define boost ios::sync_with_stdio(false);cin.tie(0)

#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)

#define max3(a,b,c) max(max(a,b), c);

#define min3(a,b,c) min(min(a,b), c);

const ll oo = 1ll<<;

const ll mos = 0x7FFFFFFF;  //

const ll nmos = 0x80000000;  //-2147483648

const int inf = 0x3f3f3f3f;

const ll inff = 0x3f3f3f3f3f3f3f3f; //

const int mod = 1e9+;

const double esp = 1e-;

const double PI=acos(-1.0);

const double PHI=0.61803399;    //黄金分割点

const double tPHI=0.38196601;

template<typename T>

inline T read(T&x){

    x=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

inline void cmax(int &x,int y){if(x<y)x=y;}

inline void cmax(ll &x,ll y){if(x<y)x=y;}

inline void cmin(int &x,int y){if(x>y)x=y;}

inline void cmin(ll &x,ll y){if(x>y)x=y;}

/*-----------------------showtime----------------------*/

            const int maxn = ;

            int pt[maxn],vis[maxn];

            struct node{

                int cnt;

                int id;

            }a[maxn];

            queue<int>que;

            vector<int>mp[maxn];

            int col = ;

            bool dfs(int u){

                if(vis[u] == col) return false;

                vis[u] = col;

                for(int i=; i<mp[u].size(); i++){

                    int v = mp[u][i];

                    if(pt[v] == || dfs(pt[v])) {

                        pt[v] = u;

                        return true;

                    };

                }

                return false;

            }

            int in[maxn];

            bool cmp(node a,node b){

                return a.cnt < b.cnt;

            }

int main(){

            int n,m,k;

            scanf("%d%d%d", &n, &m, &k);

            rep(i, , m) a[i].id = i;

            for(int i=; i<=k; i++){

                int u,v;

                scanf("%d%d", &u, &v);

                mp[v].pb(u);

                a[v].cnt++;

            }

            sort(a+, a++m, cmp);

            int ans = ;

            for(int i=; i<=m; i++) {

                   // memset(vis, 0, sizeof(vis));

                    col++;

                    if(dfs(a[i].id) == ) que.push(a[i].id);

                    else ans++;

            }

            int mx = ;

            while(!que.empty()){

                int u = que.front(); que.pop();

                for(int i=; i<mp[u].size(); i++)

                    {

                        int v = mp[u][i];

                        in[v]++;

                        mx = max(mx, in[v]);

                    }

            }

            mx = min(, mx);

            ans += mx;

            printf("%d\n", ans);

            return ;

}

G Water Testing

皮克定理

代码：

#include<bits/stdc++.h>

#define int long long

#define MAX(a,b,c) max(a,max(b,c))

#define MIN(a,b,c) min(a,min(b,c))

#define pb push_back

#define fi first

#define se second

typedef long long ll;

typedef long long LL;

typedef unsigned long long ull;

typedef unsigned long long uLL;

using namespace std;

const int maxn=1e6+;

const int mod=1e9+;

struct Point{int x;int y;}pa[maxn];

int  Cross(Point A,Point B) { return A.x*B.y-A.y*B.x;   }  // 叉积

/*bool anglecmp(Point a, Point b)

{

    if(a.y <= 0 && b.y > 0) return true;

    if(a.y > 0 && b.y <= 0) return false;

    if(!a.y && !b.y) return a.x < b.x;

    return Cross(a,b) > 0;

}*/

int work(Point a,Point b)

{

    return __gcd( abs(a.x-b.x),abs(b.y-a.y) )-;

}

int32_t main()

{

    int n; scanf("%lld",&n);

    for(int i=;i<=n;i++)  scanf("%lld %lld",&pa[i].x,&pa[i].y);

    //sort(pa+1,pa+1+n,anglecmp);

    int ans=;

    for(int i=;i<n;i++)

    ans+=Cross(pa[i],pa[i+]);

    ans+=Cross(pa[n],pa[]);

    ans=abs(ans);

    int x=n;

    for(int i=;i<n;i++)

    x+=work(pa[i],pa[i+]);

    x+=work(pa[n],pa[]);

    int a=(ans-x+)/;

    printf("%lld\n",a);

}

H Ratatoskr

枚举根，求最长链，然后取最小

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

const int N = ;

vector<int> g[N];

int dfs(int u, int o) {

    int res = ;

    for (int v : g[u]) {

        if(v != o) res = max(res, dfs(v, u)+);

    }

    return res;

}

int main() {

    int n, a, b, c, u, v;

    scanf("%d", &n);

    scanf("%d %d %d", &a, &b, &c);

    for (int i = ; i < n; ++i) scanf("%d %d", &u, &v), g[u].pb(v), g[v].pb(u);

    int ans = n;

    for (int i = ; i <= n; ++i) {

        if(i == b || i == c) ans =min(ans, dfs(i, i)-);

        else ans = min(dfs(i, i), ans);

    }

    printf("%d\n", ans);

    return ;

}

I Uberwatch

dp

代码：

#include <algorithm>

#include  <iterator>

#include  <iostream>

#include   <cstring>

#include   <cstdlib>

#include   <iomanip>

#include    <bitset>

#include    <cctype>

#include    <cstdio>

#include    <string>

#include    <vector>

#include     <stack>

#include     <cmath>

#include     <queue>

#include      <list>

#include       <map>

#include       <set>

#include   <cassert>

/*

⊂_ヽ

　 ＼＼ Λ＿Λ  来了老弟

　　 ＼('ㅅ')

　　　 >　⌒ヽ

　　　/ 　 へ＼

　　 /　　/　＼＼

　　 ﾚ　ノ　　 ヽ_つ

　　/　/

　 /　/|

　(　(ヽ

　|　|、＼

　| 丿 ＼ ⌒)

　| |　　) /

'ノ )　　Lﾉ

*/

using namespace std;

#define lson (l , mid , rt << 1)

#define rson (mid + 1 , r , rt << 1 | 1)

#define debug(x) cerr << #x << " = " << x << "\n";

#define pb push_back

#define pq priority_queue

typedef long long ll;

typedef unsigned long long ull;

//typedef __int128 bll;

typedef pair<ll ,ll > pll;

typedef pair<int ,int > pii;

typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q

//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q

#define fi first

#define se second

//#define endl '\n'

#define boost ios::sync_with_stdio(false);cin.tie(0)

#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)

#define max3(a,b,c) max(max(a,b), c);

#define min3(a,b,c) min(min(a,b), c);

const ll oo = 1ll<<;

const ll mos = 0x7FFFFFFF;  //

const ll nmos = 0x80000000;  //-2147483648

const int inf = 0x3f3f3f3f;

const ll inff = 0x3f3f3f3f3f3f3f3f; //

const int mod = 1e9+;

const double esp = 1e-;

const double PI=acos(-1.0);

const double PHI=0.61803399;    //黄金分割点

const double tPHI=0.38196601;

template<typename T>

inline T read(T&x){

    x=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

inline void cmax(int &x,int y){if(x<y)x=y;}

inline void cmax(ll &x,ll y){if(x<y)x=y;}

inline void cmin(int &x,int y){if(x>y)x=y;}

inline void cmin(ll &x,ll y){if(x>y)x=y;}

/*-----------------------showtime----------------------*/

            const int maxn = 3e5+;

            int a[maxn],dp[maxn],res[maxn];

int main(){

            int n,m;

            scanf("%d%d", &n, &m);

            rep(i, , n) scanf("%d", &a[i]);

            rep(i, m+, n) {

                dp[i] = max(dp[i],res[i-m] + a[i]);

                res[i] = max(res[i-], dp[i]);

            }

            printf("%d\n", res[n]);

            return ;

}

J Word Clock

旅行商问题变形，状压dp实现

代码:

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb push_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

const int N = ;

const int INF = 0x3f3f3f3f;

int h, w, n;

string t[N], s[N];

int d[N][N], up;

pii dp[N][<<N];

int pre[N][<<N];

pii dfs(int u, int st) {

    if(~dp[u][st].fi) return dp[u][st];

    pii res = {INF, INF};

    for (int i = ; i < n; ++i) {

        if(i == u) continue;

        if(st & (<<i)) {

            pii p = dfs(i, st^(<<u));

            p.se += d[i][u];

            if(p.se > w) p.fi++, p.se = s[u].size();

            if(p < res) {

                res = p;

                pre[u][st] = i;

            }

        }

    }

    if(res.fi == INF) res.fi = , res.se = s[u].size();

    return dp[u][st] = res;

}

int main() {

    fio;

    cin >> h >> w >> n;

    int tot = ;

    for (int i = ; i < n; ++i) cin >> t[i];

    for (int i = ; i < n; ++i) {

        bool f = true;

        if(t[i].size() > w) return *puts("impossible");

        for (int j = ; j < n; ++j) {

            if(i == j) continue;

            if(t[j].find(t[i]) == string::npos) continue;

            f = false;

        }

        if(f) s[tot++] = t[i];

    }

    n = tot;

    for (int i = ; i < n; ++i) {

        for (int j = ; j < n; ++j) {

            if(i == j) continue;

            int mn = min(s[i].size(), s[j].size());

            d[i][j] = s[j].size();

            for (int k = ; k <= mn; ++k) {

                if(s[i].substr(s[i].size()-k, k) == s[j].substr(, k)) d[i][j] = s[j].size() - k; ///!!!!!!!!!

            }

        }

    }

    up = <<n;

    for (int i = ; i < n; ++i) {

        for (int j = ; j < up; ++j)

        dp[i][j] = {-, -}, pre[i][j] = -;

    }

    for (int i = ; i < n; ++i) {

        pii p = dfs(i, up-);

        if(p.fi < h) {

            vector<string> res (h, string(w, 'X'));

            vector<int> p;

            int now = i, st = up-;

            while(~now) {

                p.pb(now);

                now = pre[now][st];

                st ^= <<p.back();

            }

            reverse(p.begin(), p.end());

            pii pos = {, };

            for (int i = ; i < p.size(); ++i) {

                int x = p[i];

                if(i == ) {

                    for (int j = ; j < s[x].size(); ++j) res[pos.fi][pos.se+j] = s[x][j];

                    pos.se += s[x].size();

                }

                else {

                    int u = p[i-];

                    if(pos.se + d[u][x] > w) {

                        pos.fi++;

                        pos.se = ;

                        for (int j = ; j < s[x].size(); ++j) res[pos.fi][pos.se+j] = s[x][j];

                        pos.se = s[x].size();

                    }

                    else {

                        int sz = s[x].size();

                        for (int j = ; j < d[u][x]; ++j) res[pos.fi][pos.se+j] = s[x][sz-d[u][x]+j];

                        pos.se += d[u][x];

                    }

                }

            }

            for (int i = ; i < h; ++i) cout << res[i] << endl;

            return ;

        }

    }

    return *puts("impossible");

}

K You Are Fired

排序

代码：

#include <algorithm>

#include  <iterator>

#include  <iostream>

#include   <cstring>

#include   <cstdlib>

#include   <iomanip>

#include    <bitset>

#include    <cctype>

#include    <cstdio>

#include    <string>

#include    <vector>

#include     <stack>

#include     <cmath>

#include     <queue>

#include      <list>

#include       <map>

#include       <set>

#include   <cassert>

/*

⊂_ヽ

　 ＼＼ Λ＿Λ  来了老弟

　　 ＼('ㅅ')

　　　 >　⌒ヽ

　　　/ 　 へ＼

　　 /　　/　＼＼

　　 ﾚ　ノ　　 ヽ_つ

　　/　/

　 /　/|

　(　(ヽ

　|　|、＼

　| 丿 ＼ ⌒)

　| |　　) /

'ノ )　　Lﾉ

*/

using namespace std;

#define lson (l , mid , rt << 1)

#define rson (mid + 1 , r , rt << 1 | 1)

#define debug(x) cerr << #x << " = " << x << "\n";

#define pb push_back

#define pq priority_queue

typedef long long ll;

typedef unsigned long long ull;

//typedef __int128 bll;

typedef pair<ll ,ll > pll;

typedef pair<int ,int > pii;

typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q

//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q

#define fi first

#define se second

//#define endl '\n'

#define boost ios::sync_with_stdio(false);cin.tie(0)

#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)

#define max3(a,b,c) max(max(a,b), c);

#define min3(a,b,c) min(min(a,b), c);

const ll oo = 1ll<<;

const ll mos = 0x7FFFFFFF;  //

const ll nmos = 0x80000000;  //-2147483648

const int inf = 0x3f3f3f3f;

const ll inff = 0x3f3f3f3f3f3f3f3f; //

const int mod = 1e9+;

const double esp = 1e-;

const double PI=acos(-1.0);

const double PHI=0.61803399;    //黄金分割点

const double tPHI=0.38196601;

template<typename T>

inline T read(T&x){

    x=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

inline void cmax(int &x,int y){if(x<y)x=y;}

inline void cmax(ll &x,ll y){if(x<y)x=y;}

inline void cmin(int &x,int y){if(x>y)x=y;}

inline void cmin(ll &x,ll y){if(x>y)x=y;}

/*-----------------------showtime----------------------*/

            const int maxn = 1e4+;

            struct node{

                string name;

                ll x;

            }a[maxn];

            bool cmp(node a, node b){

                return a.x > b.x;

            }

int main(){

            ll n, d, k;

            cin>>n>>d>>k;

            rep(i,, n)

            {

                cin>>a[i].name>>a[i].x;

            }

            sort(a+,a++n,cmp);

            ll sum = ;

            rep(i,, k) sum += a[i].x;

            if(sum < d) puts("impossible");

            else {

                printf("%d\n", k);

                rep(i, , k) {

                    cout<<a[i].name<<", YOU ARE FIRED!"<<endl;

                }

            }

            return ;

}

2017-2018 ACM-ICPC German Collegiate Programming Contest (GCPC 2017)的更多相关文章

  1. (寒假开黑gym)2017-2018 ACM-ICPC German Collegiate Programming Contest (GCPC 2017)

    layout: post title: (寒假开黑gym)2017-2018 ACM-ICPC German Collegiate Programming Contest (GCPC 2017) au ...

  2. 2017-2018 ACM-ICPC German Collegiate Programming Contest (GCPC 2017)（9/11）

    $$2017-2018\ ACM-ICPC\ German\ Collegiate\ Programming\ Contest (GCPC 2017)$$ \(A.Drawing\ Borders\) ...

  3. 2017-2018 ACM-ICPC German Collegiate Programming Contest (GCPC 2017) Solution

    A. Drawing Borders Unsolved. B. Buildings Unsolved. C. Joyride Upsolved. 题意: 在游乐园中,有n个游玩设施,有些设施之间有道路 ...

  4. 2018 German Collegiate Programming Contest (GCPC 18)

    2018 German Collegiate Programming Contest (GCPC 18) Attack on Alpha-Zet 建树,求lca 代码: #include <al ...

  5. (寒假GYM开黑)2018 German Collegiate Programming Contest (GCPC 18)

    layout: post title: 2018 German Collegiate Programming Contest (GCPC 18) author: "luowentaoaa&q ...

  6. 2015 German Collegiate Programming Contest (GCPC 15) + POI 10-T3（12/13）

    $$2015\ German\ Collegiate\ Programming\ Contest\ (GCPC 15) + POI 10-T3$$ \(A.\ Journey\ to\ Greece\ ...

  7. ACM ICPC, JUST Collegiate Programming Contest (2018) Solution

    A:Zero Array 题意:两种操作, 1 p v  将第p个位置的值改成v  2  查询最少的操作数使得所有数都变为0  操作为可以从原序列中选一个非0的数使得所有非0的数减去它,并且所有数不能 ...

  8. ACM ICPC, Amman Collegiate Programming Contest (2018) Solution

    Solution A:Careful Thief 题意:给出n个区间,每个区间的每个位置的权值都是v,然后找长度为k的区间,使得这个区间的所有位置的权值加起来最大,输出最大权值, 所有区间不重叠 思路 ...

  9. Gym .102021 .German Collegiate Programming Contest (GCPC 18) （寒假gym自训第三场）

    B .Battle Royale 题意:给你两个点A,B,以及一个圆S,保证两个点在圆外,且其连线与圆相交,求两点间最短距离. 思路:显然是要分别与圆相切,然后在圆弧想走,直到相交. 那么ans=与圆 ...

随机推荐

  1. (转)测试如何区别是前端的问题还是后台的bug

    常常说到的一个IT项目,包括前端开发,后台开发,软件测试,架构,项目经理,产品需求.那么对于一位优秀的软件测试工程师来说,需要区分前端和后台的工作就显得尤为重要. - 什么是前端和后台 简而言之,前端 ...

  2. ELK日志分析解决方案

    概要 ELK(Elasticsearch , Logstash, Kibana的简称)是目前比较流行的日志分析解决方案,核心包括了三个部分 Elasticsearch:日志查询分析引擎 Logstas ...

  3. 再次聊一聊promise settimeout asycn awiat执行顺序---js执行机制 EVENT LOOP

    首先js是单线程 分为同步和异步,异步又分为(macrotask 宏任务 和 microtask微任务 ), 这图还是很清晰嘛,再来一张 总结一下,就是遇到同步先执行同步,异步的丢到一边依次排队,先排 ...

  4. bootstrap表格添加按钮、模态框实现

    bootstrap表格添加按钮.模态框实现 原创 2017年07月20日 17:35:48 标签: bootstrap 1723 bootstrap表格添加按钮.模态框实现 - 需求: 需要表格后面每 ...

  5. 关于element-ui表格table设置header-cell-class-name样式不起作用的原因分析

    在编写表格的时候想给表头添加样式,使用 header-cell-class-name怎么都添加不上样式,检查元素发现连class都没添加上,查了很多资料有人说element之前版本不支持这属性,但我使 ...

  6. Oracle 中 编写 function 和 procedure 的注意事项

    create or replace function sum_pro( v_a number, v_b number ) return number is -- v_result number(10) ...

  7. validationEngine 使用

    引入文件 环境在 jQuery 下 , 所有先要引入 jQuery <%--校验样式--%> <link rel="stylesheet" href=" ...

  8. CentOS7.3安装zimbra8.7.11

    系统版本:CentOS7.3 Zimbra版本:8.7.11 设置IP地址 vi /etc/sysconfig/network-scripts/ifcfg-ens33 #使用vi命令修改ip BOOT ...

  9. Android SDK Manager for Mac 在线更新镜像地址截至2017-10-01亲测有效

    虽然国内google被墙了,但仍可利用国内的某些镜像网站实现Android SDK在线更新,使用方法如下: 1.启动 Android SDK Manager ,打开主界面,依次选择『Tools』.『O ...

  10. webServices接口开发过程中项目启动遇到的错误org.xml.sax.SAXParseException; lineNumber: 20; columnNumber: 422; schema_reference.4: Failed to read schema document 'http://www.springframework.org/schema/beans/spring-bean

    org.xml.sax.SAXParseException; lineNumber: 20; columnNumber: 422; schema_reference.4: Failed to read ...