Codeforces Round 731 Div3 A-G题解

Codeforces Round #731 (Div. 3)

在家打了好久COD和战雷,偶尔也得学习一下,要不然感觉时间都浪费了,游戏玩多了也腻,保持适当学习才能爽玩游戏。申请完了也不想做太难的题了,那么就来一场div3保持一下思维敏捷度吧。

A. Shortest Path with Obstacle

题解

这道题的思路很简单,判断block点和start点是否在同一行或者同一列,如果是的话,那么就需要多走2步,否则就是最短曼哈顿距离。

代码

AC代码
#include <bits/stdc++.h>
using namespace std;
constexpr int limit =  (4e5  + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}
int n, m;
int a[limit];
void solve() {
    int xa,ya;
    int xb, yb;
    cin>>xa>>ya;
    cin>>xb>>yb;
    int xf, yf;
    cin>>xf>>yf;
    int ans = abs(xa - xb) + abs(ya - yb);
    if(xa == xb){
        if(xf == xa){
            auto [l, r] = minmax(ya, yb);
            if(yf >= l and yf <= r){
                cout<<ans + 2<<endl;
                return;
            }
        }
    }
    if(ya == yb){
        if(ya == yf){
            auto [l, r] = minmax(xa, xb);
            if(xf >= l and xf <= r){
                cout<<ans + 2<<endl;
                return;
            }
        }
    }
    cout<<ans<<endl;

}
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
    int kase;
    cin>>kase;
    while (kase--)
        invoke(solve);
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s";
    return 0;
}

B. Alphabetical Strings

题解

倒着模拟就行了,没啥难度

代码

AC代码
#include <bits/stdc++.h>
using namespace std;
constexpr int limit =  (4e5  + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}
int n, m;
int a[limit];
void solve() {
    string str;
    cin>>str;
    n = str.length();
    deque<char>q{str.begin(), str.end()};
    per(i,0,n - 1){
        char c = 'a' + i;
        if(q.front() == c){
            q.pop_front();
            continue;
        }
        if(q.back() == c){
            q.pop_back();
            continue;
        }
        cout<<"NO"<<endl;
        return;
    }
    cout<<"YES"<<endl;
}
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
    int kase;
    cin>>kase;
    while (kase--)
        invoke(solve);
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s";
    return 0;
}

C. Pair Programming

题解

这个题模拟就行了,如果a不行放b,如果b不行放a,如果都不行输出-1,因为太久没写代码所以卡在了一些implementation details上面

代码

AC代码
#include <bits/stdc++.h>
using namespace std;
constexpr int limit =  (4e5  + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}
int n, m;
int a[limit];
int b[limit];
void solve() {
    int k;
    cin>>k>>n>>m;
    deque<int>p,q;
    rep(i,1,n){
        int x;
        cin>>x;
        p.push_back(x);
    }
    rep(i,1,m){
        int x;
        cin>>x;
        q.push_back(x);
    }
    vector<int>ans;
    while(p.size() or q.size()){
        if(p.size()){
            auto x = p.front();
//            cout<<x<<endl;
            p.pop_front();
            if(!x){
                ans.push_back(x);
                k++;
                continue;
            }else{
                if(x <= k){
                    ans.push_back(x);
                    continue;
                }else{
                    if(!q.size()){
                        cout<<"-1"<<endl;
                        return;
                    }
                    if(q.front() > k){
                        cout<<"-1"<<endl;
                        return;
                    }
                    ans.push_back(q.front());
                    k += not q.front();
                    q.pop_front();
                    p.push_front(x);
                }
            }
        }else{
            auto x = q.front();
//            cout<<x<<endl;
            q.pop_front();
            if(x <= k){
                ans.push_back(x);
                k += not x;
                continue;
            }else{
                if(!p.size()){
                    cout<<"-1"<<endl;
                    return;
                }
                if(p.front() > k){
                    cout<<"-1"<<endl;
                    return;
                }
                ans.push_back(p.front());
                k += not p.front();
                p.pop_front();
                q.push_front(x);
            }
        }
    }
    for(const auto &it : ans){
        cout<<it<<" ";
    }
    cout<<endl;

}
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
    int kase;
    cin>>kase;
    while (kase--)
        invoke(solve);
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s";
    return 0;
}

D. Co-growing Sequence

题解

这道题就是统计一下前一个和后一个有多少不一样的,很容易想到,没啥说的

代码

AC代码
#include <bits/stdc++.h>
using namespace std;
constexpr int limit =  (4e5  + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}
#define int ll
int n, m;
int a[limit];
void solve() {
    cin>>n;
    rep(i,1,n){
        cin>>a[i];
    }
    vector<int>ans(n + 1, 0);
    rep(i,2,n){
        ll x = a[i - 1] ^ ans[i - 1];
        ll y = a[i];
        ll res = 0;
        rep(j, 0, 30){
            ll now = 1ll << j;
            int fst = x bitand now;
            int scd = y bitand now;
            if(fst == scd or scd){
                continue;
            }
            res += now;
        }
        ans[i] = res;
    }
    for(const auto &it : ans | views::drop(1)){
        cout<<it<<" ";
    }
    cout<<endl;
}
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
    int kase;
    cin>>kase;
    while (kase--)
        invoke(solve);
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s";
    return 0;
}

E. Air Conditioners

题解

不会,没想到,等晚上回家补一下,真的不会

F. Array Stabilization (GCD version)

题解

这题刚开始看上去暴力可做,后来发现其实是没考虑需要一段的情况。我们从每一个i开始,让i和i + 1相同,就要除掉这个以i开头连续的段的所有gcd,然后所需要最长的段的长度就是我们的答案(很好理解,因为每次只能消掉一个嘛)

代码

AC代码
#include <bits/stdc++.h>
using namespace std;
constexpr int limit =  (4e5  + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}
int n, m;
int a[limit];
void solve() {
    cin>>n;
    rep(i,1,n){
        cin>>a[i];
        a[i + n] = a[i];
    }
    int ans = 0;
    rep(i,1,n){
        auto fst = a[i];
        auto scd = a[i + 1];
        int res = 0;
        int now = i + 1;
        while(fst != scd){
//            cout<<fst<<" "<<scd<<endl;
            fst = gcd(fst, a[now]);
            scd = gcd(scd,  a[now + 1]);
            ++now;
            ++res;
        }
        ans = max(ans, res);
    }
    cout<<ans<<endl;


}
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
    int kase;
    cin>>kase;
    while (kase--)
        invoke(solve);
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s";
    return 0;
}

G. How Many Paths?

题解

这题不出意外又是个大模拟

首先因为这是一个有向图,所以我们不难发现两点

  1. 如果一个点u在一个环(从点1可到达的连通分量)上,那么点u有无数条path从1到u,从u这个连通分量能够到达的点的path数也是无数的
  2. 如果一个点u不在一个环上,而从点1到点u有大于一条路径,那么从u出发的path数也是大于1的。

显然,1的优先级是大于2的,所以我们可以先找到所有的环,然后再找到所有的非环但有大于1条路径的点,然后再统计答案。

如何做呢?

首先我们要处理所有的环,把图编程connected DAG,这一步可以用tarjan来做,但是记住我们只dfs 1,如果1到不了的地方直接给0就行。然后重新建图

其次,我们对于情况2,统计从1可达的点的in degree,如果in[u] >= 2, 那么这个点就属于情况2、

最后如果不属于1或者2的,那么就属于是只有一条路径。

其实也没有看上去那么难,对吧?

代码

AC代码
#include <bits/stdc++.h>
using namespace std;
constexpr int limit =  (5e5  + 5);//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}
int n, m;
int a[limit];
vector<int>g[limit];
// tarjan 缩点
int dfn[limit], low[limit], scc[limit], cnt, scc_cnt;
stack<int> st;
void dfs(int u) {
    dfn[u] = low[u] = ++cnt;
    st.push(u);
    for (auto v : g[u]) {
        if (!dfn[v]) {
            dfs(v);
            low[u] = min(low[u], low[v]);
        } else if (!scc[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if (dfn[u] == low[u]) {
        ++scc_cnt;
        while (true) {
            int x = st.top();
            st.pop();
            scc[x] = scc_cnt;
            if (x == u)break;
        }
    }
}

vector<int>g2[limit];
int color[limit];
int in[limit]; // 入度
void solve() {
    cin>>n>>m;
    set<int>circle;
    rep(i,1,n){
        g[i].clear();
        g2[i].clear();
    }
    while(st.size())st.pop();
    fill(dfn, dfn + 1 + n, 0);
    fill(low, low + 1 + n, 0);
    fill(color, color + 1 + n, 0);
    fill(scc, scc + 1 + n, 0);
    fill(in, in + 1 + n, 0);
    rep(i,1,m){
        int x,y;
        cin>>x>>y;
        if(x == y){
            circle.insert(x);
            continue;
        } // 去除自环
        g[x].push_back(y);
    }
    scc_cnt = 0;

    cnt = 0;
    dfs(1);
    rep(i,1,n){
        if(!low[i]){
            color[i] = 0;
        }
    }
    // 接下来处理所有链路上的点
    map<int, vector<int>> grp;
    rep(i, 1, n){
        if(!low[i])
            continue;
        grp[scc[i]].push_back(i);
    }
    set<int>real_circle;
    for(const auto &[id, nodes] : grp){
        if(nodes.size() > 1 or circle.contains(nodes.front())){
            for(const auto & it : nodes){
                color[it] = -1;
            }
            real_circle.insert(id);
        }
        // 重新建图
        for(const auto &it : nodes){
            for(const auto &v : g[it]){
                g2[id].push_back(scc[v]); // 重新连边
                in[scc[v]]++;
            }
        }
    }
    int vs = scc[1];
    // 开始bfs, 先处理有环的点的连边问题
    // 此时应该是无环图
    map<int, bool>visited;
    {
        queue<int>q;
        for(auto &&src : real_circle){
            q.push(src);
            visited[src] = 1;
        }
        while(q.size()){
            auto u = q.front();
            q.pop();
            for(auto &&v : g2[u]){
                if(!visited[v]){
                    visited[v] = true;
                    q.push(v);
                }
            }
        }
    }
    for(auto &&[k, v]: visited){
        if(v){
            for(auto &&node : grp[k]){
                    color[node] = -1;
            }
        }
    }
    map<int, bool>vis;
    queue<int>q;
    rep(i,1,n){
        if(in[i] > 1){
            q.push(i);
            vis[i] = 1;
        }
    }
    while(q.size()){
        auto u = q.front();
        q.pop();
        if(visited[u])
            continue;
        for(auto &&v : g2[u]){
            if(visited[v])
                continue;
            if(!vis[v]){
                vis[v] = 1;
                q.push(v);
            }
        }
    }
    for(auto &&[id, v] : vis){
        if(visited[id])
            continue;
        for(auto &&node: grp[id]){
            color[node] = 2;
        }
    }
    rep(i,1,n){
        if(low[i]){
            if(not color[i])
                color[i] = 1;
        }
    }
    rep(i,1,n){
        cout<<color[i]<<" ";
    }
    cout<<endl;



}
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO
    int kase;
    cin>>kase;
    while (kase--)
        invoke(solve);
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s";
    return 0;
}
    
posted @ 2023-11-29 05:07  tiany7  阅读(17)  评论(0编辑  收藏  举报