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Codeforces Round 966 (Div. 3)

时间：2024-08-29 10:37:52浏览次数：11  
标签：966 int Codeforces long read while Div dis define

A. Primary Task

if-else语句练习，判断前两个字符是否为"10"，并判断之后的字符是否大于1

点击查看代码 
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define pii pair<int, int>

int read()
{
    int x = 0, f = 0; char c = getchar();
    while (c < '0' || c > '9') f = c == '-', c = getchar();
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
    return f ? -x : x;
}

const int N = 1e4 + 5;
char s[N];

int main()
{
    int T = read();
    while(T--)
    {
        scanf("%s", s + 1);
        int len = strlen(s + 1);
        if(s[1] == '1' && s[2] == '0' && ((len > 3 && s[3] >= '1') || (len == 3 && s[3] >= '2'))) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

B. Seating in a Bus

模拟

点击查看代码 
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define pii pair<int, int>

int read()
{
    int x = 0, f = 0; char c = getchar();
    while (c < '0' || c > '9') f = c == '-', c = getchar();
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
    return f ? -x : x;
}

const int N = 2e5 + 5;
int a[N], vis[N];

int main()
{
    int T = read();
    while(T--)
    {
        int n = read();
        for(int i = 1; i <= n; ++i) a[i] = read();
        for(int i = 1; i <= n; ++i) vis[i] = 0;
        vis[a[1]] = 1;
        int flag = 1;
        for(int i = 2; i <= n; ++i)
        {
            int ner = 0;
            if(a[i] > 1) ner |= vis[a[i] - 1];
            if(a[i] < n) ner |= vis[a[i] + 1];
            if(ner == 0){ flag = 0; break; }
            vis[a[i]] = 1;
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

C. Numeric String Template

还是模拟，每个数字第一次出现映射到一个字符上，之后再出现直接判断，字符第一次出现同理

点击查看代码 
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define pii pair<int, int>

int read()
{
    int x = 0, f = 0; char c = getchar();
    while (c < '0' || c > '9') f = c == '-', c = getchar();
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
    return f ? -x : x;
}

const int N = 2e5 + 5;
int a[N], b[N];
int id1[N], id2[N];
char s[N];
void solve()
{
    int n = read();
    for(int i = 1; i <= n; ++i) a[i] = b[i] = read();
    sort(b + 1, b + n + 1);
    int x = unique(b + 1, b + n + 1) - (b + 1);
    for(int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + x + 1, a[i]) - b;
    int m = read();
    while(m--)
    {
        scanf("%s", s + 1);
        int flag = 1, len = strlen(s + 1);
        if(len != n){ printf("NO\n"); continue; }
        for(int i = 1; i <= n; ++i) id1[i] = -1;
        for(int i = 1; i <= n; ++i)
        {
            if(id1[a[i]] == -1) id1[a[i]] = s[i] - 'a';
            else if(id1[a[i]] != s[i] - 'a') flag = 0;
            if(!flag) break;
        }
        for(int i = 0; i <= 25; ++i) id2[i] = -1;
        for(int i = 1; i <= len; ++i)
        {
            if(id2[s[i] - 'a'] == -1) id2[s[i] - 'a'] = a[i];
            else if(id2[s[i] - 'a'] != a[i]) flag = 0;
            if(!flag) break;
        }
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
}

int main()
{
    int T = read();
    while(T--) solve();
    return 0;
}

D. Right Left Wrong

注意到 \(LLRR\) 取 \((1, 3)(2, 4)\) 和 \((1, 4)(2, 3)\) 是等价的，于是直接取所有区间只有包含没有相交的情况，即用双指针从两端向中间扫即可

点击查看代码 
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define pii pair<int, int>

int read()
{
    int x = 0, f = 0; char c = getchar();
    while (c < '0' || c > '9') f = c == '-', c = getchar();
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
    return f ? -x : x;
}

const int N = 2e5 + 5;
int a[N];
char s[N];
ll sum[N];

void solve()
{
    int n = read();
    for(int i = 1; i <= n; ++i) a[i] = read(), sum[i] = sum[i - 1] + a[i];
    scanf("%s", s + 1);
    int l = 1, r = n;
    ll ans = 0;
    while(l < r)
    {
        while(l <= n && s[l] != 'L') ++l;
        while(r >= 1 && s[r] != 'R') --r;
        if(l < r) ans += sum[r] - sum[l - 1];
        ++l, --r;
    }
    printf("%lld\n", ans);
}

int main()
{
    int T = read();
    while(T--) solve();
    return 0;
}

E. Photoshoot for Gorillas

简单计算一下一个格子被多少个边长为 \(K\) 的子方阵包含，那么这个格子里的数就会贡献这么多次，贪心的让大数放到次数多的格子里

点击查看代码 
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define pii pair<int, int>

int read()
{
    int x = 0, f = 0; char c = getchar();
    while (c < '0' || c > '9') f = c == '-', c = getchar();
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
    return f ? -x : x;
}

const int N = 2e5 + 5;
int a[N];
int cnt[N];

bool cmp(int a, int b){ return a > b; }

void solve()
{
    int n = read(), m = read(), K = read(), W = read(), tot = 0;
    for(int i = 1; i <= W; ++i) a[i] = read();
    sort(a + 1, a + W + 1, cmp);
    for(int x = 1; x <= n; ++x)
        for(int y = 1; y <= m; ++y)
        {
            int lx = max(1, x - K + 1), rx = min(n - K + 1, x);
            int ly = max(1, y - K + 1), ry = min(m - K + 1, y);
            cnt[++tot] = 1ll * (rx - lx + 1) * (ry - ly + 1);
        }
    sort(cnt + 1, cnt + tot + 1, cmp);
    ll ans = 0;
    for(int i = 1; i <= W; ++i) ans += 1ll * a[i] * cnt[i];
    printf("%lld\n", ans);
}

int main()
{
    int T = read();
    while(T--) solve();
    return 0;
}

F. Color Rows and Columns

首先想到的是贪心，即一个 \(a \times b\) 的矩形，填满一行或一列需要 \(\min\{a, b\}\) 步

这样贪心的取，发现当 \(1 \times 1\) 时一次性填了一行和一列，答案 + 2

贪心不完全对，转而考虑背包 \(DP\)

需要注意的是：想要答案 + a + b ，需要 \(a \times b\) 步，想要答案 + a + b - 1 ，也需要 \(a \times b\) 步

点击查看代码 
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define pii pair<int, int>

int read()
{
    int x = 0, f = 0; char c = getchar();
    while (c < '0' || c > '9') f = c == '-', c = getchar();
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
    return f ? -x : x;
}

const int inf = 0x3f3f3f;
int dp[2][105], a[1005], b[1005];
vector< pair<int, int> > x[1005];

void init(int id)
{
    x[id].clear();
    int A = a[id], B = b[id], sum = 0, K = 0;
    while(A && B)
    {
        sum += A, --B, ++K;
        if(A > B) swap(A, B);
        if(A == 0) x[id].emplace_back(pair<int, int>(K, sum)), ++K;
        x[id].emplace_back(pair<int, int>(K, sum));
    }
}

void solve()
{
    int n = read(), K = read();
    for(int i = 1; i <= n; ++i)
    {
        a[i] = read(), b[i] = read();
        if(a[i] > b[i]) swap(a[i], b[i]);
        init(i);
    }
    for(int i = 1; i <= K; ++i) dp[0][i] = inf;
    int ans = inf;
    for(int i = 1, op = 1; i <= n; ++i, op ^= 1)
    {
        for(int j = 0; j <= K; ++j) dp[op][j] = dp[op ^ 1][j];
        for(auto [w, val] : x[i])
        {
            for(int j = w; j <= K; ++j)
                dp[op][j] = min(dp[op][j], dp[op ^ 1][j - w] + val);
        }
        ans = min(ans, dp[op][K]);
    }
    if(ans == inf) printf("-1\n");
    else printf("%d\n", ans);
}

int main()
{
    int T = read();
    while(T--) solve();
    return 0;
}

G. Call During the Journey

最短路魔改，松弛操作时分三种情况：

  1. 直接走路过去，不受任何限制

  2. 坐车过去，需要保证下车时间 \(\le t1\) 或者上车时间 \(\ge t2\)

  3. 等到 \(t2\) 再上车

点击查看代码 
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define pii pair<int, int>

int read()
{
    int x = 0, f = 0; char c = getchar();
    while (c < '0' || c > '9') f = c == '-', c = getchar();
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
    return f ? -x : x;
}

const ll inf = 0x7fffffffffff;
const int N = 1e5 + 5;  
int n, m;
int t0, t1, t2;

struct node
{
    int v, l1, l2;
    node(){ v = l1 = l2 = 0; }
    node(int x, int y, int z){ v = x, l1 = y, l2 = z; }
};

vector<node> e[N];

ll dis[N];
bool vis[N];
priority_queue< pair<int, int> > q;

bool check(int t)
{
    // printf("t = %d\n", t);
    for(int i = 1; i <= n; ++i) dis[i] = inf, vis[i] = 0;
    dis[1] = t;
    q.push(pair<int, int>(-dis[1], 1));
    while(!q.empty())
    {
        int u = q.top().second ;
        q.pop();
        if(vis[u]) continue;
        vis[u] = 1;
        // printf("u = %d, dis = %lld\n", u, dis[u]);
        for(auto [v, l1, l2] : e[u])
        {
            int flag = 0;
            if(dis[v] > dis[u] + l2)
            {
                dis[v] = dis[u] + l2;
                flag = 1;
            }
            if(dis[v] > dis[u] + l1 && ((dis[u] + l1 <= t1) || (dis[u] >= t2)))
            {
                dis[v] = dis[u] + l1;
                flag = 1;
            }
            if(dis[u] <= t2 && dis[v] > t2 + l1)
            {
                dis[v] = t2 + l1;
                flag = 1;
            }
            if(flag) q.push(pair<int, int>(-dis[v], v));
        }
    }
    if(dis[n] <= t0) return true;
    else return false;
}

void solve()
{
    n = read(), m = read();
    t0 = read(), t1 = read(), t2 = read();
    for(int i = 1; i <= n; ++i) e[i].clear();
    for(int i = 1; i <= m; ++i)
    {
        int u = read(), v = read(), l1 = read(), l2 = read();
        e[u].emplace_back(node(v, l1, l2));
        e[v].emplace_back(node(u, l1, l2));
    }
    int l = -1, r = t0;
    while(l < r)
    {
        int mid = (l + r + 1) >> 1;
        if(check(mid)) l = mid;
        else r = mid - 1;
    }
    printf("%d\n", l);
}

int main()
{
    int T = read();
    while(T--) solve();
    return 0;
}

H. Ksyusha and the Loaded Set

转化题意为，最初所有数都为 \(1\) ，集合中包含的数为 \(0\) ，三种操作：

  1. 加入集合一个数，等价于将此位置设为 \(0\)

  2. 删去集合一个数，等价于将此位置设为 \(1\)

  3. 询问集合中1的连续段长度 \(\ge k\) 的最小左端点

权值线段树搞就完了

终于知道多测的意义了，专门卡 \(O(值域)\) 的算法

注意到每次测试时都重建一颗一模一样的树，时间消耗主要在这里，不优

又注意到每个测试中的操作数是 \(O(n)\) 的，考虑撤销操作

只在一开始建树，之后每次用时撤销前面的测试的操作

总结

  • 权值线段树遇上多测，考虑撤销操作
点击查看代码 
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define ull unsigned long long
#define pii pair<int, int>

int read()
{
    int x = 0, f = 0; char c = getchar();
    while (c < '0' || c > '9') f = c == '-', c = getchar();
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
    return f ? -x : x;
}

const int N = 2e5 + 5;
int n, m, a[N];

struct node
{
    int mx, lsum, rsum, len;
    node(){ mx = lsum = rsum = len = 0; }
    node friend operator + (node a, node b)
    {
        node c;
        c.mx = max(a.mx , b.mx );
        c.mx = max(c.mx , a.rsum + b.lsum );
        c.len = a.len + b.len ;
        c.lsum = a.lsum + ((a.lsum == a.len ) ? b.lsum : 0);
        c.rsum = b.rsum + ((b.rsum == b.len ) ? a.rsum : 0);
        return c;
    }
};

#define ls(x) (x << 1)
#define rs(x) (x << 1 | 1)

node A[8388605 + 10];

void build(int k, int l, int r)
{
    if(l == r)
    {
        A[k].mx = A[k].lsum = A[k].rsum = A[k].len = 1;
        return ;
    }
    int mid = (l + r) >> 1;
    build(ls(k), l, mid), build(rs(k), mid + 1, r);
    A[k] = A[ls(k)] + A[rs(k)];
}

void update(int k, int l, int r, int pos, int val)
{
    if(l == r)
    {
        A[k].mx = A[k].lsum = A[k].rsum = val;
        return ;
    }
    int mid = (l + r) >> 1;
    if(pos <= mid) update(ls(k), l, mid, pos, val);
    else update(rs(k), mid + 1, r, pos, val);
    A[k] = A[ls(k)] + A[rs(k)];
}

int query(int k, int l, int r, int len)
{
    if(l == r) return l;
    int mid = (l + r) >> 1;
    if(A[ls(k)].mx >= len) return query(ls(k), l, mid, len);
    if(A[ls(k)].rsum + A[rs(k)].lsum >= len) return mid - A[ls(k)].rsum + 1;
    return query(rs(k), mid + 1, r, len);
}

pair<int, int> sta[N << 1];
int top;

void solve()
{
    n = read();
    for(int i = 1; i <= n; ++i) a[i] = read();
    m = read();
    top = 0;
    for(int i = 1; i <= n; ++i) update(1, 1, 4e6, a[i], 0), sta[++top] = pair<int, int>(a[i], 0);
    while(m--)
    {
        char c;
        int x;
        scanf(" %c %d", &c, &x);
        if(c == '-') update(1, 1, 4e6, x, 1), sta[++top] = pair<int, int>(x, 1);
        if(c == '+') update(1, 1, 4e6, x, 0), sta[++top] = pair<int, int>(x, 0);
        if(c == '?') printf("%d ", query(1, 1, 4e6, x));
    }
    while(top)
    {
        update(1, 1, 4e6, sta[top].first , sta[top].second ^ 1);
        --top;   
    }
    printf("\n");
}

int main()
{
    int T = read();
    build(1, 1, 4e6);
    while(T--) solve();
    return 0;
}

标签：966,int,Codeforces,long,read,while,Div,dis,define
From： https://www.cnblogs.com/wenqizhi/p/18386129

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