codeforces-1133 (div3)

时间：2023-03-08 19:01:55

A.先全部化成分钟数，取平均数之后化成正常时刻。

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int main(){

    int h1,m1,h2,m2;

    scanf("%d:%d %d:%d",&h1,&m1,&h2,&m2);

    int num = (h2 - h1) *  + (m2 - m1);

    num /= ; m1 += num;

    while(m1 >= ){

        h1++;

        m1 -= ;

    }

    printf("%02d:%02d",h1,m1);

    return ;

}

A

B.所有数全部对K取余之后就能知道可以匹配他的兄弟是谁了，计数直接算即可。

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int a[maxn];

int num[maxn];

int main(){

    Sca2(N,K);

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

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

        a[i] %= K;

        num[a[i]]++;

    }

    int ans = num[] / ;

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

        if(i + i == K) ans += num[i] / ;

        else ans += min(num[i],num[K - i]);

    }

    Pri(ans * );

    return ;

}

B

C.一个类似简化版的尺取即可。

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int a[maxn];

int main(){

    Sca(N);

    for(int i = ; i <= N ; i ++) Sca(a[i]);

    sort(a + ,a +  + N);

    int s = ,t = ;

    int ans = ;

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

        while(t < N && a[t + ] - a[i] <= ) t++;

        ans = max(ans,t - i + );

    }

    Pri(ans);

    return ;

}

C

D.由于使每个位置取0的D有一个固定的数，map统计一下相同D的最大个数即可。

这里选择用结构体重载一下运算符处理，double 和 long double 都WA了，所以化成了分数最简比。

要注意如果要用map去重，不但要重载 == ,还有 <,最好可以每一个元素都在 < 里面讨论到，使得map排序结果可以确定

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const long double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

LL a[maxn],b[maxn];

struct Node{

    LL x,y;

    Node(){}

    Node(LL x,LL y):x(x),y(y){}

    friend bool operator == (Node a,Node b){

        return a.x == b.x && a.y == b.y;

    }

    friend bool operator < (Node a,Node b){

        if(a.x == b.x) return a.y < b.y;

        return a.x < b.x;

    }

};

LL gcd(LL a,LL b){

    return !b?a:gcd(b,a % b);

}

map<Node,int>P;

int main(){

    Sca(N);

    for(int i = ; i <= N ; i ++) scanf("%lld",&a[i]);

    for(int i = ; i <= N ; i ++) scanf("%lld",&b[i]);

    int ans = ,flag = ;

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

        if(!a[i]){

            if(!b[i]) flag++;

            continue;

        }

        Node d;

        if(!b[i]){

            d = Node(,);

        }else{

            LL g = gcd(abs(a[i]),abs(b[i]));

            int flag = ;

            if(a[i] < ) flag *= -;

            if(b[i] < ) flag *= -;

            d = Node(flag * abs(b[i]) / g,abs(a[i]) / g);

        }

        P[d]++;

        ans = max(ans,P[d]);

    }

    Pri(ans + flag);

    return ;

}

D

E.C题的扩展，dp[i][j]记录有1 ~ j之间选i只队最多可以选多少人，直接递推即可。

5000 * 5000有很大MLE的风险，但是第一层i可以直接省略掉，空间复杂度为O(n)

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int a[maxn];

int pre[maxn];

int dp[maxn];

int main(){

    Sca2(N,K);

    for(int i = ; i <= N ; i ++) Sca(a[i]);

    sort(a + ,a +  + N);

    int t = ;

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

        while(t <= N && a[t + ] - a[i] <= ){

            t++; pre[t] = i;

        }

    }

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

        for(int j = N; j >= ; j --){

            dp[j] = max(dp[j],dp[pre[j] - ] + j - pre[j] + );

        }

        for(int j = ; j <= N ; j ++) dp[j] = max(dp[j],dp[j - ]);

    }

    Pri(dp[N]);

    return ;

}

E

F1.取ind最多的点，将所有和他相邻的点率先加入树边，然后并查集处理加入余下的树边

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int ind[maxn];

int fa[maxn];

void init(){

    for(int i =  ; i <= N ; i ++) fa[i] = i;

}

int find(int t){

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

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

}

int Union(int a,int b){

    a = find(a); b = find(b);

    if(a == b) return ;

    fa[a] = b;

    return ;

}

PII E[maxn];

int main(){

    Sca2(N,M); init();

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

        int u,v; Sca2(u,v);

        E[i].fi = u; E[i].se = v;

        ind[u]++; ind[v]++;

    }

    int Max = ;

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

        if(ind[Max] < ind[i]) Max = i;

    }

    int ans = ;

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

        if(E[i].fi == Max || E[i].se == Max){

            if(Union(E[i].fi,E[i].se)){

                ans++;

                printf("%d %d\n",E[i].fi,E[i].se);

            }

        }

    }

    for(int i = ; i <= M && ans < N - ; i ++){

        if(Union(E[i].fi,E[i].se)){

            ans++;

            printf("%d %d\n",E[i].fi,E[i].se);

        }

    }

    return ;

}

F1

F2.一开始以为点1出去的桥必须联通，其他任意，后来发现并不。

删除所有1相邻的边之后产生的连通分量，1需要有边和每一个连通分量有边，其余剩下的位置才是任意。

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

struct Edge{

    int to,next;

    bool cut;

}edge[maxn * ];

int head[maxn],tot;

int fa[maxn];

int sum;

void init(){

    for(int i =  ; i <= N ; i ++) head[i] = -;

    tot = ;

}

int find(int x){

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

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

}

int Union(int a,int b){

    a = find(a); b = find(b);

    if(a == b) return ;

    fa[a] = b; sum++;

    return ;

}

void add(int u,int v){

    edge[tot].to = v;

    edge[tot].next = head[u];

    edge[tot].cut = ;

    head[u] = tot++;

}

int ind[maxn];

vector<PII>ans;

int main(){

    Sca3(N,M,K); init();

    for(int i =  ; i <= N ; i ++) fa[i] = i;

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

        int u = read(),v = read();

        add(u,v); add(v,u);

        if(u !=  && v != ) Union(u,v);

        ind[u]++; ind[v]++;

    }

    for(int j = head[]; ~j ; j = edge[j].next){

        if(Union(,edge[j].to)){

            edge[j].cut = edge[j ^ ].cut = ;

        }

    }

    sum = ;

    for(int i = ; i <= N ; i ++) fa[i] = i;

    for(int i = head[]; ~i; i = edge[i].next){

        if(edge[i].cut && Union(,edge[i].to)){

            ans.pb(mp(,edge[i].to));

            K--;

        }

    }

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

        for(int j = head[i]; ~j; j = edge[j].next){

            if((i ==  || edge[j].to == ) && K <= ) continue;

            if(Union(edge[j].to,i)){

                if(i ==  || edge[j].to == ) K--;

                ans.pb(mp(i,edge[j].to));

            }

        }

    }

    if(sum < N -  || K){

        puts("NO");

        if(M == ){

            cout << sum << " " << K << endl;

        }

        return ;

    }

    puts("YES");

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

        printf("%d %d\n",ans[i].fi,ans[i].se);

    }

    return ;

}

F2

相关文章