【AtCoder】ARC076

时间  2019-12-09
标签 AtCoder arc076 arc 繁體版
原文   原文链接

ARC076

C - Reconciled?

若是\(N = M\)node

答案是\(2N!M!\)c++

若是\(|N - M| = 1\)ui

答案是\(N!M!\)spa

不然答案是0code

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,M;
int fac[MAXN];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void Solve() {
    read(N);read(M);
    if(abs(N - M) > 1) {puts("0");return;}
    fac[0] = 1;
    for(int i = 1 ; i <= max(N,M) ; ++i) {
    fac[i] = mul(fac[i - 1],i);
    }

    int ans = mul(fac[N],fac[M]);
    if(N == M) {
    ans = mul(ans,2);
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - Built?

发现横坐标排序和纵坐标排序后，只有相邻的边会有边排序

而后跑最小生成树便可ci

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N;
int x[MAXN],y[MAXN];
int id[MAXN],cnt,fa[MAXN];
struct node {
    int u,v,c;
}E[MAXN * 10];
int getfa(int u) {
    return fa[u] == u ? u : fa[u] = getfa(fa[u]);
}
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {read(x[i]);read(y[i]);}
    for(int i = 1 ; i <= N ; ++i) id[i] = i;
    sort(id + 1,id + N + 1,[](int a,int b) {return x[a] < x[b];});
    for(int i = 1 ; i < N ; ++i) {
    E[++cnt] = (node){id[i],id[i + 1],x[id[i + 1]] - x[id[i]]};
    }
    for(int i = 1 ; i <= N ; ++i) id[i] = i;
    sort(id + 1,id + N + 1,[](int a,int b) {return y[a] < y[b];});
    for(int i = 1 ; i < N ; ++i) {
    E[++cnt] = (node){id[i],id[i + 1],y[id[i + 1]] - y[id[i]]};
    }
    sort(E + 1,E + cnt + 1,[](node a,node b){return a.c < b.c;});
    for(int i = 1 ; i <= N ; ++i) fa[i] = i;
    int ans = 0;
    for(int i = 1 ; i <= cnt ; ++i) {
    if(getfa(E[i].u) != getfa(E[i].v)) {
        ans += E[i].c;
        fa[getfa(E[i].u)] = getfa(E[i].v);
    }
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Connected?

只有两个点都在边上的连线才有影响，咱们把这些点按照顺时针（逆时针也能够）扔进栈里，栈顶元素和它相同则弹出，若是最后栈是空的就合法，不然就不合法get

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
struct node {
    int x,y,p;
};
int R,C;
int N;
int x[MAXN][2],y[MAXN][2];
vector<node> v[4];
int sta[MAXN * 2],top;
bool on_u(int x,int y) {
    return x == 0;
}
bool on_l(int x,int y) {
    return x != 0 && x != R && y == 0;
}
bool on_r(int x,int y) {
    return x != 0 && x != R && y == C;
}
bool on_d(int x,int y) {
    return x == R;
}
bool check(int x,int y) {
    return on_u(x,y) || on_l(x,y) || on_r(x,y) || on_d(x,y);
}
int id(int x,int y) {
    if(on_u(x,y)) return 0;
    else if(on_r(x,y)) return 1;
    else if(on_d(x,y)) return 2;
    else return 3;
}
void Solve() {
    read(R);read(C);read(N);
    for(int i = 1 ; i <= N ; ++i) {
    for(int j = 0 ; j <= 1 ; ++j) {read(x[i][j]);read(y[i][j]);}
    if(check(x[i][0],y[i][0]) && check(x[i][1],y[i][1])) {
        v[id(x[i][0],y[i][0])].pb((node){x[i][0],y[i][0],i});
        v[id(x[i][1],y[i][1])].pb((node){x[i][1],y[i][1],i});
    }
    }
    sort(v[0].begin(),v[0].end(),[](node a,node b){return a.y < b.y;});
    sort(v[1].begin(),v[1].end(),[](node a,node b){return a.x < b.x;});
    sort(v[2].begin(),v[2].end(),[](node a,node b){return a.y > b.y;});
    sort(v[3].begin(),v[3].end(),[](node a,node b){return a.x > b.x;});
    for(int i = 0 ; i < 4 ; ++i) {
    for(auto t : v[i]) {
        if(sta[top] == t.p) --top;
        else sta[++top] = t.p;
    }
    }
    if(top) puts("NO");
    else puts("YES");
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Exhausted?

线段树维护hall定理，好像以前遇到过相似的可是没有写这个作法。。。it

hall定理是什么呢class

\(X,Y\)为两边的点集，\(X\)和\(Y\)之间存在完美匹配，仅当\(X\)中任意\(k\)个点组成的点集和\(Y\)中的至少$k
$个点相连

这是充分必要的

那么咱们求的是人数-最大匹配，也就是每一个点集减去匹配\(Y\)点集椅子个数的最大值，初始的答案先设成\(max(N - M,0)\)

设点集大小为\(|X|\)，匹配\(Y\)点集的椅子数是\(\Gamma(X)\)

答案显然不小于\((|X| - |\Gamma(X)|)_{max}\)

若是答案大于\((|X| - |\Gamma(X)|)_{max}\)，设\(t > (|X| - |\Gamma(X)|)_{max}\)，若是当前存在一个大于t的未匹配点，那么选这t个点必定与对面至少一个点相连，设为\(v\)，若v被匹配，找到匹配点u，u与这t个点组成的点集匹配至少两个点，因此能够增长一个匹配点，能够证实答案不大于\((|X| - |\Gamma(X)|)_{max}\)

因此只要考虑怎么求\((|X| - |\Gamma(X)|)_{max}\)

显然是对于一个固定的区间，计算多少个点的\(L_{i},R_{i}\)覆盖了这个区间，因此把区间按左端点排序，而后线段树维护便可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 +c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,M;
struct node {
    int l,r;
}p[MAXN];
struct seg_tree {
    int l,r;
    int lz,mx;
}tr[MAXN * 4];
void update(int u) {
    tr[u].mx = max(tr[u << 1 | 1].mx,tr[u << 1].mx);
}
void addlz(int u,int v) {
    tr[u].mx += v;tr[u].lz += v;
}
void pushdown(int u) {
    if(tr[u].lz) {
    addlz(u << 1,tr[u].lz);
    addlz(u << 1 | 1,tr[u].lz);
    tr[u].lz = 0;
    }
}
void build(int u,int l,int r) {
    tr[u].l = l;tr[u].r = r;
    if(l == r) {
    tr[u].mx = r;
    return;
    }
    int mid = (l + r) >> 1;
    build(u << 1,l,mid);
    build(u << 1 | 1,mid + 1,r);
    update(u);
}
void Change(int u,int l,int r,int v) {
    if(tr[u].l == l && tr[u].r == r) {addlz(u,v);return;}
    pushdown(u);
    int mid = (tr[u].l + tr[u].r) >> 1;
    if(r <= mid) Change(u << 1,l,r,v);
    else if(l > mid) Change(u << 1 | 1,l,r,v);
    else {Change(u << 1,l,mid,v);Change(u << 1 | 1,mid + 1,r,v);}
    update(u);
}
int Query(int u,int l,int r) {
    if(tr[u].l == l && tr[u].r == r) {return tr[u].mx;}
    pushdown(u);
    int mid = (tr[u].l + tr[u].r) >> 1;
    if(r <= mid) return Query(u << 1,l,r);
    else if(l > mid) return Query(u << 1 | 1,l,r);
    else return max(Query(u << 1,l,mid),Query(u << 1 | 1,mid + 1,r));
}
void Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= N ; ++i) {
    read(p[i].l);read(p[i].r);
    }
    build(1,1,M + 1);
    sort(p + 1,p + N + 1,[](node a,node b){return a.l < b.l || (a.l == b.l && a.r > b.r);});
    int ans = max(0,N - M);
    for(int i = 1 ; i <= N ; ++i) {
    Change(1,p[i].l + 1,p[i].r,1);
    int t = Query(1,p[i].l + 1,M + 1) - M - p[i].l - 1;
    ans = max(ans,t);
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}
相关文章
相关标签/搜索
每日一句
    每一个你不满意的现在，都有一个你没有努力的曾经。
本站公众号
   欢迎关注本站公众号,获取更多信息
联系我们 最近搜索 最新文章 沪ICP备13005482号-10 MyBatis教程 SQL 教程 MySQL教程 Java 教程 Thymeleaf 教程 Hibernate教程 Spring教程 Redis教程